GIFT   OF 


WER  DlVISldff 


TEACHERS'    MANUAL 


TO   ACCOMPANY 


A  TEXT-BOOK  IN  PHYSICS 


WILLIAM  X.   MUMPER,    Ph.D. 

•ROFE: 

MODEL  SCHOOL-  :SEY 


\niXXATI  •:-  0  If  1C  A  <  JO 

AMERICAN    BOOK    COMPANY 


UNIVERSITY  OF   CALIFORNIA 


LIBRARY 

OF  THE 


Received.. 
Accessions  No. 


Book  No. /. 

r*/6 


LOWER  DIVISION 


TEACHERS'    MANUAL 


TO  ACCOMPANY 


A  TEXT=BOOK  IN  PHYSICS 


BY 


WILLIAM  N.  MUMPER,  Ph.D. 

PROFESSOR    OF    PHYSICS    IN    THE    STATE    NORMAL    AND 
MODEL    SCHOOLS    OF    TRENTON,    NEW    JERSEY 


NEW  YORK  - :  •  CINCINNATI  • :  •  CHICAGO 

AMERICAN    BOOK   COMPANY 


M  7 


NOTE 

Slight  disagreements  between  the  numerical  answers  to 
some  of  the  problems  as  here  given  and  those  obtained  by 
others  will  probably  in  all  cases  be  due  to  the  fact  that  many 
of  the  physical  quantities  used  are,  of  necessity,  approximate 
values,  and  to  the  additional  fact  that  some  of  the  solutions 
were  obtained  by  means  of  logarithms. 


COPYRIGHT,  1909,  BY  WILLIAM  N.  MUMPER. 
w.  P.    i 


;  f\ 


TEACHERS'  MANUAL  FOR  MUMPER'S 
PHYSICS 

INTRODUCTORY— Page  19 

1.  The  capacity  may  be  found  by  counting  the  number  of  gallons 
required  to  fill  it.    This  is  a  direct  measurement.    It  may  also  be  found 
by  making  certain  linear  measurements  and  from  these  computing  the 
volume.     This  is  an  indirect  measurement. 

2.  1  m.=  1000  mm.,  3  m.  =  3000  mm. 
1  mm.  =  .001  m.,  35  mm.  =  .035  m. 
1  m.=  100  cm.,  48  m.  =  4800  cm. 

1  in.  =  2.54  cm.,  10  in.  =  25.4  cm. 

1  cm.  =  .3937  in. 

25  cm.  =  25X.3937  in.  =  9.8+  in. 

3.  1  k.  =  2.21b.,  12  k.=  12X2.21b.  =  26.41b. 
1  oz.  =  28.3+  gm. 

4.  1  ft.  =  30.48  cm.,  J^l  =  32.1  +  ft.  per  sec. 

5.  It  is  not  correct.     The  second  statement  means  that  when  the 
age  of  the  tree  becomes  any  multiple  of  its  present  age  the  height  will 
become  the  same  multiple  of  its  present  height. 

MATTER— Page  25 

1.  Its  number  of  units  of  volume  and  its  number  of  units  of  mass. 

2.  The  mass  may  be  found  by  weighing.    See  page  24  (1).    The  vol- 
ume of  a  block  of  wood  may  be  found  by  measuring  the  volume  of  the  liq- 
uid it  displaces  when  immersed.    The  surface  of  the  wood  should  be  made 
impervious  by  a  coat  of  paraffin  or  varnish. 

The  volume  of  a  liquid  may  be  found  by  measuring  it  in  a  graduated 
vessel.  Its  mass  may  be  found  by  weighing. 

3.  The  volume  of  the  liquid  is  measured  directly  and  the  other  quan- 
tities are  measured  indirectly. 

4.  See  Table,  page  25. 

6.  1  c.  c.  of  water  weighs  1  gm.,  1  liter=1000  c.  c.,  hence  1  1.  of  water 
weighs  1000  gm.,  45  c.  c.  of  water  weigh  45  gm. 

6.  1  c.  c.  is  merely  a  volume  and  a  volume  has  no  weight.  Objects 
which  have  a  volume  of  1  c.  c.  may  have  different  weights  depending 
upon  their  densities. 

M^.MUM.PH.S.-l       32^825 


'   MANUAL  FOE  MUMPER'S  PHYSICS 

7.  If  32  grii.  niass  fill  66  :c:  c.,  the  density  is  32-^-60=  .53+  gm.  per  c.  c. 

8.  See  page  16,  par.  9.     The  kilogram  mass  is  the  same  everywhere 
though  the  earth's  attraction  for  it,  that  is,  its  weight  may  change  with  a 
change  of  location. 

9.  The  assumption  is  that  at  a  given  place  equal  masses  have  equal 
weights,  that  is,  they  are  attracted  equally  by  the  earth. 

10.  The  volume  is  5X6X7X1  c.  c.  =  210  c.  c. 
The  density  is  1638 -=-210=  7.8  gm.  per  c.  c. 

11.  12  c.  c.  holds  12X  13.6  gm.=  163.2  gm.  of  mercury. 

1  liter  of  mercury  contains  1000X13.6  gm.=  13600  gm. 
1  k.=  1000  gm.,  1000-1-13.6=73.5  c.  c. 

12.  Volume  of  tank=  2X3X4X1  cu.  ft.  =  24  cu.  ft. 
1  cu.  ft.  of  water  weighs  62.4  Ib. 

24  cu.  ft.  of  water  weigh  24X62.4  lb.  =  1497.6. 

1  cu.  ft.  of  mercury  weighs  848  Ib. 

24  cu.  ft.  weigh  24X848  Ib.  =  20352  Ib. 

MATTER— Page  38 

1.  By  showing  that  air  occupies  space,  has  weight,  elasticity  and 
other  characteristics  common  to  all  matter. 

2.  The  gas  molecules  occupy  a  part  of  the  space  between  the  molecules 
of  the  liquid. 

3.  Let  the  pupil  take  different  substances  and  demonstrate  by  his 
experiments  and  his  descriptions  of  them  that  he  knows  the  proper  terms 
to  use  in  describing  his  observations. 

4.  The  surface  tension  of  the  liquid  wax  makes  it  approach  as  nearly 
as  possible  to  the  spherical  shape.    See  page  35,  line  6. 

6.  The  spaces  between  the  fibers  of  the  towel  act  like  capillary  tubes, 
or  like  the  glass  plates.    See  Fig.  22. 
6.  Essentially  the  same  as  5. 

THE  MECHANICS  OF  LIQUIDS— Page  43 

1.  The  intensity  is  64-f-16=4  Ib.  per  sq.  in.     External  pressure,  be- 
cause produced  by  a  body  acting  on  the  fluid. 

2.  The  intensity   of   external   pressure   is  the  same  throughout  the 
fluid;  it  is  18  gm.  per  sq.  cm. 

Area  is  8X12X1  sq.  cm.  =  96  sq.  cm. 

The  total  pressure  is  96X18  gm.=  1728  gm. 

3.  The  intensities  are  equal. 

The  total  pressures  and  the  areas  are  directly  proportional;  hence  the 
areas  are  to  each  other  as  3:16. 

4.  7200  fb. -^30=  240  Ib.  on  1  sq.  in.;  being  external  pressure  it  is  also 
240  Ib.  per  sq.  in.  on  the  small  piston. 


THE  MECHANICS  OF  LIQUIDS  3 

5.  40  ft>.  on  2.5  in.    The  intensity  is  40-1-2.5=  16.  Ib.  per  sq.  in.  (Pas- 
cal's law). 

6.  A  total  pressure  of  2000  Ib.  is  to  be  produced  on  the  piston.     The 
intensity  is  40  Ib.  on  1  sq.  in.    2000 H-  40=  50  sq.  in.  to  produce  2000  Ib. 
To  lift  the  elevator  the  pressure  must  exceed  the  computed  amount.    Just 
how  much  greater  depends  upon  friction,  speed  desired,  etc. 

THE  MECHANICS  OF  LIQUIDS— Pages  53,  54 

1.  Laws  1  and  2,  page  48.    Page  48,  last  two  lines.    20  ft.  deep  the 
pressure  intensity  is  20X62.4  ft>.=  1248  Ib.  per  sq.  ft.     When  water  is 
1  cm.  deep  the  pressure  is  1  gm.  per  sq.  cm.,  hence  for  18  cm.  deep  it  is 
18X1  gm.  =  18  gm.  per  sq.  cm. 

2.  See  page  49,  line  10  from  bottom.     Since  the  intensity  is  13.6  gm. 
per  sq.  cm.  when  the  mercury  is  1  cm.  deep,  when  it  is  7  cm.  deep  the 
intensity  is  7X13.6  gm.  =  95.2  gm.  per  sq.  cm.     When  alcohol  is  1  cm. 
deep  the  intensity  pressure  is  .79  gm.  per  sq.  cm.    When  4  cm.  deep  it 
presses  4X.79  gm.  =  3.16  gm.  per  sq.  cm. 

3.  We  must  use  an  average  depth  when  the  water  does  not  have  the 
same  depth  at  all  points  of  the  surface  in  question.     The  average  depth 
is  here  \  (0,  least  depth  +  40  cm.,  greatest  depth)  =  20  cm.     Hence  the 
average  pressure  intensity  is  20  gm.  per  sq.  cm.     When  containing  mer- 
cury, average  depth  is  20  cm.,  pressure  intensity =20 XI 3. 6  gm.  =  272  gm. 
per  sq.  cm. 

4.  The  area  of  the  bottom  is  224  sq.  cm. 

(a)    Pressure  intensity  is  12  X  1  gm.=  12  gm.  per  sq.  cm. 

Total  pressure  is  224X12  gm.=  2688  gm. 
(6)    Pressure  intensity  is  12 X  13.6  gm.=  163.2  gm.  per  sq.  cm. 

Total  pressure  is  224X163.2  gm.  =  36556.8  gm. 

5.  Area  of  1  side,  12X16X1  sq.  cm. =  192  sq.  cm. 
Average   depth    is    £   (0+12)  =  6 

cm. 
(a)    Average  pressure  intensity  is  6  gm. 

per  sq.  cm. 
Total  pressure    is   192X6    gm.= 

1152  gm. 

(6)    Average  depth  is  6  cm.  and  aver- 
age pressure  intensity  is   6X 

13.6   gm.  =  81.6   gm.   per   sq. 

cm. 
Total    pressure     is     192X81.6= 

15667.2  gm. 

6.  Area  of  the  surface  which  is  exposed  to  the  pressure  of  the  water  on 
only  one  side  is  10X20X1  sq.  ft.  =  200  sq.  ft. 


4  MANUAL  FOR  MUMPER'S  PHYSICS 

The  average  depth  on  this  area  is  5  ft. 

The  intensity  of  pressure  is  5X62.4=312  Ib.  per  sq.  ft. 

The  total  excess  of  pressure  is  200X312  =  62400  Ib. 

7.  (a)  When  pistons  are  on  same    level,   only  external  pressure  is 
concerned  and  the  pressure  intensities  are  the  same  at  both  pistons. 

(6)  The  weight  pressure  of  the  liquid  is  added  to  the  external  pressure 
at  the  large  piston,  hence  the  total  pressure  intensity  is  greater  than 
that  at  the  smaller  piston,  and  the  denser  the  liquid  used  the  greater  the 
difference  between  the  intensities  at  the  two  pistons. 

8.  Since  the  depth  at  all  points  of  the  bottom  is  16  cm.  the  pressure 
intensity  is  16  gm.  per  sq.  cm. 

The  area  of  the  bottom  is  80  sq.  cm.,  hence  the  total  pressure  on 

the  bottom  is  80X16  gm.=  1280  gm. 
The  area  of  side  abed  is  160  sq.  cm.,  the  average  depth  is  \  (OX  16)  = 

8  cm.,  and  the  average  pressure  intensity  is  8  gm.  per  sq.  cm. 
Total  pressure  on  this  side  is  160X8  gm.=  1280  gm. 
Since  the  free  surface  is  on  the  level  of  the  top  of  the  vessel  there 

is  no  weight  pressure  there. 

9.  (a)  When  the  tube  also  is  full  of  water  the  depth  at  the  level  adfe 
is  20  cm.  and  the  depth  at  all  points  of  the  bottom  is  36  cm.,  hence  the 
average  depth  on  all  points  of  the  side  abed  is  £  (20 +36)  =  28  cm. 

(6)  The  average  pressure  intensity  on  side  abed  is  28  gm.  per  sq.  cm. 
The  area  of  abed  is  160  sq.  cm.,  hence  the  total  pressure  on  abed  is  160 X 
28gm.  =  4480gm. 

10.  A  change  in  diameter  of  the  tube  has  no  effect  when  depth  and 
density  of  liquid  remain  constant. 

A  change  of  liquid  would  change  the  results  because  the  density  is 
different — in  this  case  less  than  that  of  water. 

11.  (a)  The  pressure  intensity  is  12  gm.  per  sq.  cm.  on  the  bottom 
of  each. 

(6)  Since  the  area  of  the  bottom  is  in  each  case  100  sq.  cm.  the  total 
pressure  in  each  case  is  100X12  gm.=  1200  gm. 

(c)  Since  sides  of  A  are  vertical,  total  pressure  on  the  bottom  of  A  is 
equal  to  the  weight. 

(d)  In  vessels  C  and  F. 

(e)  In  vessels  B,  D,  and  probably  E. 

12.  When  the  water  is  at  rest,  the  vertical  depth  of  the  water  below 
the  free  surface  of  the  water  in  the  reservoir.     The  same  for  a  pipe  on 
second  floor.     The  pressure  will  be  less  than  the  values  thus  computed 
in  case  the  water  is  running,  the  decrease  in  pressure  being  caused  by 
friction. 

13.  The  head  at  any  time  is  the  vertical  depth  of  water  required  to 
produce  the  pressure  of  the  water  at  any  point  at  the  given  time. 

14.  If  the  pressure  at  any  point  is  5  times  as  great  as  that  at  another 


THE  MECHANICS  OF  GASES  AND  LIQUIDS  5 

point  in  the  same  body  of  water  the  depth  is  5  times  as  great,  hence  the 
depth  of  the  reservoir  is  5X4.5  ft.=  22.5  ft. 

THE  MECHANICS  OF  GASES  AND  LIQUIDS— Pages  58,  59 

1.  Air  is  so  highly  compressible  that  a  given  motion  of  the  first  piston 
may  not  produce  any  motion  of  the  other.    There  is  lost  motion. 

2.  If  the  cap  is  on,  air  cannot  enter  there  and  the  atmospheric  pressure 
at  the  outlet  keeps  the  oil  in.     An  intermittent  flow  is  due  to  the  occa- 
sional entrance  of  air  bubbles  at  the  inlet. 

3.  The  air  pressure  at  A  plus  the  weight  pressure  of  the  liquid  above 
the  level  of  0  equals  in  intensity  the  atmospheric  pressure  at  0.    As  the 
ink  is  used  the  level  at  0  will  be  lowered  till  an  air  bubble  enters  the  bulb 
through  the  horizontal  tube  and  an  equal  volume  of  ink  flows  to  the 
outlet  side. 

4.  The  air  pressure  at  b  (Fig.  56)  and  at  A  (Fig.  57)  are  each  less 
than  the  atmospheric  pressure.    In  Fig.  58  the  air  pressure  is  greater  than 
the  atmospheric  pressure. 

5.  Since  the  water  is  10  cm.  deeper  in  column  b  than  it  is  in  column  a 
and  the  atmospheric  pressure  at  a  balances  the  air  pressure  in  b  plus  the 
pressure  of  this  depth  of  water,  the  excess  is  10  gm.  per  sq.  cm.     When 
mercury  is  used  the  excess  is  10 X  13.6  gm.=  136  gm.  per  sq.  cm. 

6.  In  charging  the  filler  the  atmospheric  pressure  drives  in  the  ink  as 
the  rubber  top  expands.    The  action  of  the  lungs  decreases  the  pressure 
within  the  straw  and  the  atmospheric  pressure  sends  the  liquid  up  the 
straw. 

7.  In  breathing,  the  chest  cavity  is  enlarged  and  the  atmospheric 
pressure  drives  the  air  into  the  lungs.    The  air  is  not  pulled  or  drawn  in 
as  is  often  assumed. 

8.  The  water  stops  rising  when  the  pressure  at  B  plus  the  pressure 
of  the  water  above  the  level  of  A  equals  the  atmospheric  pressure.     The 
atmospheric  pressure  is  6  gm.  per  sq.  cm.  greater  than  the  air  pressure  at  B. 

9.  As  the  tumbler  goes  deeper  the  water  pressure  becomes  greater, 
thus  compressing  the  air. 

10.  When  the  finger  is  removed  the  atmospheric  pressures  at  A  and 
0  are  practically  equal  and  the  weight  pressure  of  the  water  makes  it 
flow  out.  When  the  top  is  closed  there  is  no  atmospheric  pressure  at  A, 
and  the  atmospheric  pressure  at  0  is  greater  than  the  weight  pressure  of 
the  water,  hence  the  flow  stops. 

THE  MECHANICS  OF  GASES  AND  LIQUIDS— Page  64 

1.  When  mercury  is  1  cm.  deep  the  pressure  intensity  is  13.6  gm. 
per  sq.  cm.  WTien  20  cm.  deep  it  is  20X13.6  gm.=  272  gm.  per  sq.  cm. 
When  76  cm.  deep  it  is  76X  13.6  gm.=  1033.6  gm.  per  sq.  cm. 


6  MANUAL  FOR  MUMPER'S  PHYSIOS 

2.  This  means  that  the  column  of  mercury  (ab,  Fig.  61)  is  76  cm.  deep, 
hence  the  pressure  intensity  is  1033.6  gm.  per  sq.  cm.   (last  problem). 
To  compute  a  total  pressure  the  area  must  be  stated. 

3.  In  the  barometer  the  atmospheric  pressure  counterbalances  the 
weight  pressure  of  the  mercury.     It  has  been  shown  that  intensity  of 
weight  pressure  does  not  depend  upon  either  the  quantity  of  the  liquid 
or  the  shape  of  the  vessel.     Capillary  action  in  a  glass  tube  containing 
mercury  gives  depression,  hence  the  actual  reading  is  less  than  the  true 
reading.    It  is  the  reverse  in  a  water  and  glass  barometer. 

4.  Since  mercury  is  13.6  times  as  dense  as  water  a  water  column  must 
be  13.6  times  as  deep  as  the  mercury  column  to  produce  the  same  pres- 
sure—13.6X30  in.  =  408  in.  =  34  ft.     13.6X25  in.  =  340  in.  =  28^  ft. 

5.  Mercury  1  ft.  deep  exerts  a  pressure  of  13.6X62.4  lb.  =  848.6  Ib. 
per  sq.  ft.     When  1  inch  deep  the  pressure  would  be  ^  of  848.6  lb.= 
70.7+lb.  per  sq.  ft.    When  30  in.  deep  this  pressure  is  30X70.7  lb.  =  2121 
ft>.  per  sq.  ft.  or  2121 -=-144=  14.7  Ib.  per  sq.  in. 

6.  Pressure  intensity  of  mercury  is  13.6  gm.  per  sq.  cm.  for  each  cm. 
of  depth.    To  produce  980  gm.  per  sq.  cm.  the  depth  must  be  980-M3.6= 
72.0  cm. 

7.  The  same  as  the  height  of  the  standard  barometer,  76  cm.     To 
produce  the  same  pressure  water  must  be  13.6X76.  cm. =  1033. 6  cm. 

8.  Air  would,  if  uniformly  dense,  exert  a  pressure  of  .001293  gm.  for 
each  cm.  of  depth,  hence  to  produce  1033.  gm.  per  sq.  cm.  it  would  have 
to  be  1033 -f-. 00 1293  =  7989 18  cm.  =  7989  meters=5  miles  nearly. 

9.  See  use  of  the  barometer,  page  62. 

THE  MECHANICS  OF  GASES  AND  LIQUIDS— Pages  66,  67 

1.  The  weight  pressure  of  the  water  at  480  cm.  depth  is  480  gm.  per 
sq.  cm.    The  atmospheric  pressure  when  the  barometer  is  70  cm.  is  70 X 
13.6  gm.  =  952  gm.  per  sq.  cm.     The  pressure  due  to  both  is  480  gm.+ 
952  gm.=  1432  gm.  per  sq.  cm. 

2.  At  10  m.  or  1000  cm.  depth  the  weight  pressure  of  the  water  is 
1000  gm.  per  sq.  cm.  and  the  standard  atmospheric  pressure  transmitted 
to  it  is  1033.  gm.  per  sq.  cm.,  hence  the  12  c.  c.  of  air  is  under  a  pressure 
of  1000  +  1033=2033  gm.  per  sq.  cm. 

At  the  surface  the  bubble  is  under  the  atmospheric  pressure   only. 
Let  z=the  volume  at  the  surface.    Then,  according  to  Boyle's  law, 
1033:2033=  12:  re.     z=23.6c.c. 

3.  Since  the  pressure  of  the  remaining  air  is  equal  to  that  of  9  cm.  of 
mercury  and  the  atmospheric  pressure  is  72  cm.  of  mercury,  the  pressure 
of  the  remaining  air  is  9X13.6  gm.=  122.4  gm.  per  sq.  cm.;  79^  of  the 
original  mass  of  air  remains  and  ff=£  of  the  air  was  removed. 

4.  A  difference  of  1  cm.  in  level  of  mercury  surfaces  gives  a  pressure 


THE  MECHANICS  OF  GASES  AND  LIQUIDS  7 

of  13.6  gm.  per  sq.  cm.    To  produce  200  gm.  per  sq.  cm.  the  difference 
in  level  must  be  200-M3.6=14.7  cm. 

6.  The  animal  enlarges  the  chest  cavity  and  the  external  pressure 
fills  the  lungs.  In  exhalation  the  animal  reduces  the  volume  of  the  chest 
cavity  and  lungs,  driving  out  the  air. 

6.  As  the  air  leaves  the  bell  jar  (B,  Fig.  72)  the  pressure  of  the  remain- 
ing air  becomes  less  than  that  of  the  air  in  A,  which  as  it  expands  drives 
the  water  over  into  vessel  0. 

7.  If  the  mercury  has  fallen  \  of  the  distance  which  represents  the 
whole  pressure,  according  to  Boyle's  law,  the  remaining  air  which  exerts  f 
of  the  pressure  must  be  £  of  the  original  mass. 

THE  MECHANICS  OF  GASES  AND  LIQUIDS— Pages  71,  72 

1.  An  immersed  body  is  pressed  upon  by  the  fluid  in  all  directions, 
perpendicular  to  the  different  surfaces  of  the  body.     The  pressure  in- 
tensity is  the  same  at  all  points  on  the  same  level,  but  it  is  greater  at  that 
one  of  two  points  where  the  depth  is  greater,  consequently  the  pressure 
intensity  is  greatest  at  the  lowest  points  of  the  immersed  body. 

2.  Since  the  greatest  pressure  is  on  the  lowest  points  and  the  pressure 
there  is  upward,  it  follows  that  the  upward  pressure  of  a  fluid  on  an  im- 
mersed body  always  exceeds  the  downward  pressure  on  the  same  body, 
hence  a  part  or  all  of  the  weight  of  the  body  is  supported  by  the  fluid. 
This  excess  of  upward  pressure  is  called  buoyancy. 

3.  1  cu.  ft.  of  water  weighs  62.4 -|-  Ib.    Its  weight  is  supported  by  the 
cu.  ft.  of  water  or  the  surface  of  the  tank  beneath  it.    The  upward  pressure 
on  this  cu.  ft.  must  exceed  the  downward  just  enough  to  sustain  the 
weight  of  1  cu.  ft.  of  water,  62.4  Ib. 

4.  A  cubic  foot  of  any  other  material  immersed  in  the  water  would 
experience  the  same  pressure  as  does  the  cu.  ft.  of  water.     The  upward 
exceeds  the  downward  pressure  by  just  62.4  Ib.,  the  weight  of  the  water 
displaced.    This  is  less  than  the  weight  of  1  cu.  ft.  of  iron  or  of  marble, 
hence  they  would  sink;  but  it  is  greater  than  the  weight  of  1  cu.  ft.  of 
wood  and  1  cu.  ft.  of  oil,  hence  they  would  rise  in  the  water. 

5.  (a)  3X4X5X1  c.  c.  =  60  c.  c.,  volume  of  the  block. 

(6)  When  completely  immersed  it  displaces  60  c.  c.  or  (since  1  c.  c.  of 
water  weighs  1  gm.)  it  displaces  60  gm.  of  water. 

(c)  The  buoyancy  is  equal  to  the  weight  of  water  displaced,  60  gm. 

6.  (a)  450  gm.— 60  gm.  =  390  gm.  unsupported  weight.     The  block 
sinks. 

(6)  When  released  in  mercury  it  displaces  60  c.  c.  of  mercury  which 
weigh  60X13.6  gm.  =  8l6  gm.  The  buoyancy  is  816  gm.  and  the  ob- 
ject weighs  450  gm.,  hence  the  buoyancy  exceeds  the  weight  (816  —  450) 
and  the  block  rises. 


8  MANUAL  FOR  MUMPER'S  PHYSICS 

7.  When  the  body  has  the  same  density  as  that  of  the  fluid,  it  dis- 
places its  own  weight.    When  it  is  denser  it  displaces  less  than  and  when 
less  dense  it  displaces  more  than  its  own  weight  of  the  fluid  in  which  it  is 
immersed. 

8.  Since  a  floating  body  always  displaces  its  own  weight  of  the  fluid 
and  this  body  displaces  320  c.  c.  of  water,  it  must  displace  320  gm.  of 
water,  and  the  body  weighs  320  gm. 

9.  Air  has  weight  and  the  presence  of  air  in  the  water-tight  com- 
partments would  add  to  the  total  weight  of  the  boat  that  much.    Hence 
the  boat  could  carry  a  slightly  greater  weight  if   these  air-tight  com- 
partments were  empty. 

10.  (a)  A  balloon  rises  when  its  entire  weight  is  less  than  that  of  the 
air  it  displaces. 

(6)  It  falls  when  the  balloon's  weight  exceeds  that  of  the  displaced  air. 
(c)   It  will  remain  at  a  given  elevation  when  the  two  weights  are  equal. 

11.  Throwing  out  sand  decreases  the  weight  of  the  balloon  without 
materially  decreasing  the  weight  of  displaced  air.    Letting  out  some  gas 
decreases  the  volume  of  the  balloon  and  consequently  decreases  the 
buoyancy  without  materially  decreasing  the  weight  of  the  balloon.     Be- 
cause the  sand  is  several  thousand  times  as  dense  as  the  gas,  a  loss  of  sand 
changes  the  weight  of  the  balloon  without  affecting  buoyancy  appreciably, 
but  a  loss  of  gas  decreases  the  weight  very  little  and  the  buoyancy  very 
much. 

12.  Net  displacement  here  means  that  the  weight  of  the  vessel  when 
equipped  for  sea  is  20000  tons,  not  including  the  cargo.     The  displace- 
ment is  now  equal  to  the  weight  of  the  ship  plus  the  weight  of  the  cargo. 

13.  A  body  rises  when  it  displaces  more  than  its  own  weight  of  the  air. 
Particles  of  dust  and  smoke  are  solid  particles  and  they  rarely,  if  ever, 
displace  as  much  as  their  own  weight  of  air.    They  generally  fall — very 
slowly — through  the  air  in  which  they  are.     If  the  air  is  rising  it  carries 
these  particles  with  it. 

14.  A  body  is  attracted  by  the  earth  as  much  in  both  cases,  but  when 
it  is  surrounded  by  air  (or  by  any  other  fluid)  a  part  of  the  weight  is  sus- 
tained by  this  air  and  the  apparent  weight  is  less. 

15.  When  placed  in  a  vacuum  they  will  not  balance,  because  the  cork 
on  account  of  its  larger  volume  experienced  a  greater  loss  of  weight  when 
they  were  in  the  air  and  their  apparent  weights  balanced.    In  the  vacuum 
the  cork  weighs  more  than  does  the  iron. 

16.  The  true  weight  of  the  cotton  is  greater,  and  it  has  a  greater  mass 
than  that  of  the  iron. 

17.  Weight  pressure  depends  upon  depth  and  density  of  the  fluid  only. 
Hence  if  the  boat  is  of  nearly  the  same  size  and  shape  as  the  dry  dock 
a  very  small  quantity  of  water  will  produce  the  required  depth. 

18.  The  pressure  compresses  the  air  in  both  a'  and  a.     When  the  air 


THE  MECHANICS  OF  GASES  AND  LIQUIDS  9 

is  compressed  at  a  the  object  displaces  less  than  its  own  weight  of  water, 
hence  it  sinks.  When  the  pressure  of  the  finger  is  released  the  air  at  a 
expands  and  the  weight  of  water  displaced  exceeds  the  weight  of  the 
diver,  hence  it  rises. 

THE  MECHANICS  OF  GASES  AND  LIQUIDS— Pages  76,  77 

1.  It  means  that  the  weight  of  any  piece  of  copper  is  8.8  times  as 
much  as  the  weight  of  an  equal  volume  of  water.    1  c.  c.  of  copper  weighs 
8.8  gm.    In  water  it  weighs  1  gm.  less,  the  weight  of  the  1  c.  c.  of  water 
it  displaces. 

2.  10  c.  c.  of  water  weigh  10  gm.  14.8-:- 10=  1.48,  the  specific  gravity  of 
chloroform. 

3.  1  cu.  ft.  of  water  weighs  62.4  lb.=  58.8 -^62.4  =.94,  the  specific 
gravity  of  linseed  oil. 

1  c.  c.  of  water  weighs  1  gm.  and  because  the  linseed  oil  is  .94  times 
as  dense  as  the  water,  1  c.  c.  of  it  will  weigh  .94  gm. 

4.  904  gm.  — 791  gm.=  113  gm.,  the  weight  of  an  equal  volume  of 
water.    904-^-113= 8.  sp.  gr.  of  the  brass. 

(See  Archimedes'  principle.) 

6.   (a)  The  coal  weighs  1.5  times  as  much  as  an  equal  volume  of  water, 
hence  1  cu.  ft.  of  coal  will  weigh  in  air  1.5X62.4  lb.  =  93.6  Ib. 
(6)  In  water  it  will  weigh  93.6  lb.-62.4  lb.  =  31.2  Ib. 

6.  Let  x  =  loss  of  weight  in  water.    Then  1.5  =100 -hz.    z=100-M.5= 
66.6+  Ib.    100—66.6=  33.3  Ib.,  the  weight  of  the  coal  in  water. 

7.  When  floating  a  piece  of  ice  displaces  its  own  weight  of  water.    In 
fresh  water  .9  of  the  volume  of  the  ice  will  be  immersed,  for  a  volume  of 
water  .9  as  large  as  the  ice  will  weigh  just  as  much  as  the  ice.    In  salt 
water  a  smaller  portion  will  be  immersed.    .9 -h  1.025=  .88—.    Hence  (a)  .1 
of  the  volume  is  above  fresh  water  and  (6)  .12+  of  the  volume  is  above 
the  surface  of  salt  water. 

8.  1  c.  c.  of  cast  iron  (sp.  gr.  7.4)  weighs  7.4  gm.  and  10  c.  c.  weigh 
74.  gm.     In  gasoline  it  will  displace  10  c.  c.,  or   10X-66  gm.  =  6.6  gm. 
It  will  weigh  in  gasoline  74—6.6=67.4  gm. 

9.  Since  1  c.  c.  of  mercury  weighs  13.6  gm.,  the  volume  of  300  gm. 
will  be  300-M3.6=22+c.  c. 

10.  The  body  displaces  8  c.  c.  or  8  gm.  of  water.    Its  weight  in  air  is 
54.4  gm.+8  gm.  =  62.4  gm.    Its  sp.  gr.  is  62.4^8=7.8. 

11.  1   liter=1000   c.  c.     1000   c.  c.  of  cork  weigh  1000X.24  gm.= 
240  gm.;  and  it  will  displace  240  gm.  of  water  when  floating.    240  c.  c.  of 
water  will  be  displaced.     240-MOOO=  .24,  or  24%. 

12.  170  gm.— 150  gm.  =  20  gm.,  weight  of  water  displaced. 
170  gm. — 152.6=  17.4,  weight  of  turpentine  displaced. 
Volume  of  nickel  is  20  c.  c. 

(a)   Sp.  gr.  of  nickel  is  170^20=8.5. 


10  MANUAL  FOR  MUMPER'S  PHYSICS 

(6)  The  nickel  displaces  20  c.  c.  of  turpentine  which  weigh  17.4  gm. 
but  an  equal  volume  of  water  weighs  20  gm.  Sp.  gr.  of  turpentine 
is  17.4-^-20  =.87. 

13.  100  c.  c.  of  cast  iron  weigh  100X7.4  gm.  =  740  gm.    Since  the  iron 
floats  it  displaces  its  own  weight,  740  gm.  of  mercury. 

14.  The  block  of  iron  now  displaces  both  water  and  mercury,  the  joint 
weights  of  which  are  equal  to  the  weight  of  the  iron,  740  gm.     Hence  the 
block  of  iron  rises  a  little  when  the  water  is  poured  in.     The  volume  of 
water  and  mercury  displaced  are   together  equal  to  that  of   the  iron, 
100  c.  c. 

Let  x=vol.  of  mercury;  then  100 — x=vol.  of  water  displaced. 

1  c.  c.  of  mercury  weighs  13.6  gm: 

13.6  x=no.  of  gm.  of  mercury  displaced. 

100—  x  =  no.  of  gm.  of  water  displaced. 

13.6  x  + 100— x  =  total  weight  of  iron,  740  gm. 

12.6  x=  740— 100. 

re  =50. 7  c.  c.,  the  volume  of  mercury  displaced. 
100— z= 49. 3  c.  c.,  the  volume  of  water  displaced. 
Floating  in  mercury  alone  the  block  displaces  740-7-13.6=54.4  c.  c. 
54.4  c.  c. — 50.7  c.  c.  =  3.7  c.  c.,  less  mercury  displaced  after  water  is 
in. 

15.  40   gm.— 20    gm.  =  20    gm.,    weight    of   alcohol   displaced.      Since 
1  c.  c.  of  alcohol  weighs  .79  gm.,  20  gm.  weight  have  a  vol.  of  20-r-. 79= 
25.3  c.  c.,  hence  the  same  vol.  of  water  as  that  of  sugar  weighs  25.3  gm.    Sp. 
gr.  of  sugar=  40 -=-25.3=  1.58. 

64 

16.  64_i-85_33==-55  sp.  gr.  of  the  wood. 

17.  156  gm.— 73  gm.  =83  gm.,  weight  of  water  required  to  fill  bottle. 
148  gm.  — 73  gm.  =  75  gm.,  weight  of  oil  required  to  fill  bottle. 
Sp.  gr.  of  oil=75-r-S3=.9+. 

18.  (a)  The  boat  displaces  its  own  weight  or  the  weight  of  580  cu.  ft. 
of  water,  which  is  580X62.4  lb.=  36192  Ib.    It  displaces  the  same  weight 
of  sea  water  the  volume  of  which  is  580-:- 1.025  cu.  ft.  =  565.8  cu.  ft.    It  will 
be  nearer  the  deck  when  in  fresh  water,  for  it  requires  a  larger  volume  of 
the  fresh  water  to  equal  the  weight  of  the  boat. 

19.  The  vessel  will  hold  an  equal  volume  of  mercury,  which  weighs 
13.6X68  gm.  =  924.8  gm.     The  volume  of  mercury  equals  the  volume 
of  water,  which  is  68  c.  c. 

20.  Since  1  c.  c.  of  mercury  weighs  13.6  gm.,  the  volume  of  mercury 
is  here  480-f-13.6=  35.3  c.  c.    The  same  volume  of  alcohol  weighs  35.3 X  .79 
gm.  =  27.8  gm. 

21.  When   the   pressures   of  two   liquid   columns  counterbalance,  the 
relative  densities  or  specific  gravities  of  the  two  liquids  are  inversely 
proportional  to  the  depths  of  the  counterbalancing  columns. 


THE  MECHANICS  OF  GASES  AND  LIQUIDS  11 

Sp.  gr.  of  water:  sp.  gr.  of  benzine = depth  of  benzine:  depth  of  water, 
1:. 9=27:  x. 

x=. 9X27  =  24.3  cm.,  the  depth  of  the  water. 

22.  20%  water  means  1  part  water  to  4  parts  alcohol.    The  weight  of 
5  parts,  say  5  c.  c.,  is  1  gm.  +  (4X-79  gm.)  =  4.l6  gm.     An  equal  vol.  of 
water  weighs  5  gm.    The  sp.  gr.  is  4.16-^-5=  .83. 
Let  z  =  percentage  of  alcohol, 
then  100— x = percentage  of  water. 
Each  c.  c.  will  weigh  (zX-79  gm.)  +  (100  — z)  =  .88  gm. 
.79x-x=  -.12. 
-.21x=-.  12. 
21z=12. 
z=.571. 

THE  MECHANICS  OF  GASES  AND  LIQUIDS— Pages  86,  87 

1.  The  intensities  are  equal.    The  air  originally  in  the  receiver  divides 
itself  between  the  receiver  and  the  cylinder  in  the  ratio  of  their  relative 
capacities,  3:1.    |  comes  out  and  f  remains.    On  the  second  double  stroke 
£  of  the  first  remainder  comes  out  and  f  remains.     That  is,  £  of  f=y3g- 
comes  out,  and  f  of  f  or  T\  of  the  original  mass  remains. 

2.  When  the  man  who  moves  the  wheelbarrow  precedes  it  he  is  said 
to  pull,  when  he  follows  he  is  said  to  push.    Air  and  other  fluids  have  so 
little  cohesion  that  they  cannot  be  pulled  or  drawn  in  a  column,  the  object 
responsible  for  their  motion  must  follow,  hence  push  them. 

3.  The  air  pressure  must  equal  the  water  pressure  at  the  given  depth. 
At  1  m.  (100  cm.)  the  water  pressure  is  100  gm.  per  sq.  cm.,  at  15  m.  deep 
the  pressure  is  1500  gm.  per  sq.  cm.     The  atmospheric  pressure  is  75 X 
13.6  gm.=  1020  gm.  per  sq.  cm.    Since  the  air  in  the  bell  counterbalances 
both,  its  pressure  is  1500+1020=2520  gm.  per  sq.  cm. 

For  1  ft.  depth  the  pressure  is  62.4  Ib.  per  sq.  ft. 

For    50    ft.    depth   the  pressure    is   50X62.4    lb.  =  3120  Ib.   per 

sq.  ft.  water  pressure. 
Mercury  1  ft.  deep  exerts  a  pressure  of  13.6X62.4  =  848  Ib.  per 

sq.  ft. 
30  in.  =  ff=f  ft.    -|X848   lb.  =  2120   Ib.   per   sq.   ft.   atmospheric 

pressure.     3120+2120=5240  Ib.  per  sq.  ft.  due  to  both. 

4.  Since  the  water  rises  1  cm.  for  each  gram  per  sq.  cm.,  it  will  rise 
1000  cm.  or  10  m.  when  the  pressure  is  1000  gm.  per  sq.  cm. 

5.  When  the  barometer  is  80  cm.  the  atmospheric  pressure  is  80 X 
13.6  gm.=  1088  gm.  per  sq.  cm.    The  water  will  rise  1088  cm.=  10.88  m. 
Kerosene,  which  is  .8  times  as  dense  as  water,  will  rise  1088  -f-  .8=  1360  cm. 
=  13.6m. 

6.  The  pressure  is  70X13.6  gm.  =  952  gm.  per  sq.  cm.,  hence  water 


12  MANUAL  FOR  MUMPER'S  PHYSICS 

would  rise  952  cm.    And  a  liquid  1.5  times  as  dense  would  rise  9524- 1.5= 
634.6  cm. 

7.  Since  intensity  of  weight  pressure  depends  only  upon  the  depth 
and  density  of  the  liquid,  it  will  work  just  as  well  as  though  AT  had  a 
diameter  the  same  as  OT. 

8.  A  pressure  of  13  Ib.  per  sq.  in.  is  equal  to  a  pressure  of  1872  Ib. 
per  sq.  ft.    A  column  of  water  1  ft.  high  has  a  weight  pressure  of  62.4  Ib. 
per  sq.  ft.     Since  1872-^62.4=30,  an  atmospheric  pressure  of  13  Ib.  per 
sq.  in.  will  support  a  column  of  water  30  ft.  high. 

9.  When  o  is  closed  with  the  finger  and  air  is  removed  from  b  by 
means  of  a  pump  or  the  mouth,  the  atmospheric  pressure  at  a  drives 
the  liquid  over  the  high  point  T  and  fills  the  siphon. 

MOTION— Page  91 

1.  See  pages  88  and  89. 

2.  See  page  89,  subtopic,  "Motion  and  Rest  are  Relative." 

3.  Generally  speaking  there  is  little  or  no  relative  motion.     While 
turning  from  one  street  into  another  there  is  relative  motion,  which  here 
consists  in  a  change  of  direction  from  each  other  but  not  a  change  in 
distance. 

MOTION— Pages  94,  95 

1.  The  train  is  capable  of  the  greater  speed,  but  the  horse,  for  a  time 
at  least,  can  gain  speed  more  rapidly.    The  horse  is  capable  of  the  greater 
acceleration. 

2.  10  ft.  per  sec.  gained  in  1  sec.  is  the  same  rate  of  gain  or  accelera- 
tion as  is  600  ft.  per  sec.  gained  in  1  min.    The  first  car  gains  speed  60  times 
as  fast  as  the  second  car  gains  it;  its  acceleration  is  60  times  as  great. 

3.  1  mi.  per  min.  =  5280  ft.  per  min. 
1  mi.  in  3  min.=  1760  ft.  per  min. 

The  speed  of  the  train  is  3  times  that  of  the  horse. 
Ace.  of  train  5280 -f- 120  =44  ft.  per  min.  in  each  second. 
Ace.  of  horse  1760-:- 10=  176  ft.  per.  min.  in  each  second. 
The  horse  has  4  times  the  acceleration  of  the  train,  but  the  train  gains 
more  speed  because  it  keeps  gaining  for  a  longer  time. 

4.  At  the  end  of  the  first  second  the  speed  is  2  ft.  per  sec. 
At  the  end  of  10  seconds  the  speed  is  20  ft.  per  sec. 

At  the  end  of  1  min.  (60  sec.)  the  speed  is  120  ft.  per  sec. 
6.  It  will  require  30-1-2=15  sec. 

1  mi.  or  5280  ft.  per  min.  =  5280 -=-60  =88  ft.  per  sec. 

To  gain  speed  of  88  ft.  per  sec.  it  will  require  88-1-2=44  sec. 
6.  The  speed  at  the  end  of  10  sec.  is  20  ft.  per  sec.,  at  the  beginning 


MOTION  13 

it  is  0;  hence  the  average  speed  is  £  (0+20)  or  10  ft.  per  sec.    It  will  travel 
10 X 10  ft.=  100  ft.  in  10  sec. 

7.  It  means  that  a  falling  body  gains  a  velocity  of  980  ft.  per  sec. 
for  each  second  of  its  fall. 

8.  A  rising  body  loses  speed  at  the  same  rate  that  a  falling  body  gains 
it.     32.16  ft.  per  sec.  in  each  sec.  or  980.2  cm.  per  sec.  for  each  sec.  at 
New  York. 

9.  In  7  sec.  it  gains  a  velocity  of  7X980.2=6861.4  cm.  per  sec.    In 
the  seventh  second  it  gains  as  much  velocity  as  it  does  in  any  other 
second. 

10.  d=$  at  2  =  98(X22X(5)3= 490.1X25=  12252.5  cm.,  distance  fallen  in 
5  sees. 

d  =(980.2 -f-  2)  X  36  =17643.6  cm.,  dist.  in  6  sec. 
17643.6—12252.5=5391.1  cm.,  dist.  fallen  in  6th  second. 

11.  In  first  7  seconds  d=  490.1X49=  24014.9  cm. 
In  first  8  seconds  d=  490.1X64=  31366.4  cm. 
31366.4-24014.9=7351.5  cm.,  distance  fallen  in  eighth  second. 

12.  In  1  sec.  the  gain  is  32.-}-  ft.  per  sec.,  to  gain  112  ft.  per  sec.  it  will 
require  112-^32=3^  sec.  nearly.    In  1  sec.  the  gain  is  980+  cm.  per  sec.; 
to  gain  4410  cm.  per  sec.  it  will  require  4410-r-980=4.5  sec. 

13.  (a)  It  will  lose  32.  +  ft.  per  sec.  in  each  second,  hence  it  will  stop 
in  320^-32=10  seconds. 

(6)    It  rises  as  far  as  a  body  falls  in  10  sec.    d=  32X 100-:- 2=  1600  ft. 

(c)  It  will  acquire  the  same  velocity  as  it  had  when  starting  upward, 
320  ft.  per  sec. 

(d)  It  will  require  the  same  time,  10  sec. 

14.  30  mi.  per  hour  =  30-f-  3600  =T^  of  a  mile  per  sec.  =  5280 -M  20= 
44  ft.  per  sec.    To  stop  the  car  or  remove  all  its  velocity  in  10  sec.;  in 
1  sec.  the  velocity  must  be  decreased  -fa  of  44  ft.  per  sec.,  which  is  4.4  ft. 
per  sec.    To  stop  it  in  4  sec.  the  velocity  must  be  decreased  at  the  rate  of 
\  of  44=  11  ft.  per  sec.  in  each  sec. 

15.  The  distance  =£  (4.4X102)  =  220  ft.  when  stopped  in  10  sec.    Dis- 
tance =  \  (11 X  (4)2)  =  88  ft.  when  time  is  4  sec. 

16.  At  end  of  1  sec.  the  velocity  V=i+at=  10+32=42  ft.  per  sec. 
At  end  of  8  sec.  V=  10 +8(32)  =  266  ft.  per  sec. 

At  end  of  2  sec.  V=  10+2(32)  =  74  ft.  per  sec. 

At  the  beginning  V  is  10  ft.  per  sec. 

Hence  average  V  =  %  (10+74)  =  42  ft.  per  sec. 

D  =  it+$aP.     Distance=(10X8)  +  (16X82)  =  1104  ft. 

17.  When  the  weight  is  greater  the  mass  to  be  moved  is  also  greater; 
for  example,  if  the  mass  of  the  second  body  is  2  times  that  of  the  first  its 
weight  or  the  earth  pull  is  also  twice  as  great.    With  double  the  mass  we 
have  double  the  force,  hence  the  same  acceleration. 


14  MANUAL  FOR  MUMPER'S  PHYSICS 

18.  980.2  cm.   per  sec.   at  New  York.     Less  at  Washington  and  at 
Panama.    More  at  Boston  and  at  Quebec. 

19.  The  weight  of  a  body  depends  upon  its  mass,  and  the  intensity  of 
the  earth's  attraction  at  the  place  where  the  body  is.     The  mass  means 
how  much  there  is  of  it  and  the  acceleration  measures  the  attraction  in- 
tensity.    Hence  the  weight  varies  as  mass X  ace. 

20.  (a)   F  =  z+crf=20  +  (12XlO)  =  140  cm.  per  sec. 
(6)    D  =  it -H  at  ^  =  20  +  ^  -  26  cm. 

(c)  D=(20XlO)+i  (12X(10)2)=800  cm. 

(d)  Dist.  for  10  sec.  — dist.  for  9  sec.  =  dist.  for  tenth  sec. 
D  for  10  sec.=  (20X10)-K¥X(10)2)=800cm. 

D  for  9  sec.  =  (20  X  9)  +  ( V2-  X  (9)2)  =  666  cm. 
800—666=134  cm.,  dist.  passed  over  in  tenth  second. 

MOTION— Page  101 

1.  (1)  It  is  equal  to  and  opposite  to  the  force  on  the  breech. 

(2)  The  powder   acts  for  the 
same  time  on  each. 

(3)  The   directions   are    oppo- 
site. 

FIG.  2  (4)  The  velocity  of  the  bullet 

is  160  times  the  velocity  of  the  recoil   of   the  gun,   each  being  free  to 
move. 

(5)    160  (mass  of  gun) X  1=160  (speed  of  bullet) XI. 

2.  The  rapidly  flowing  water  when  suddenly  checked  exerts  a  pressure 
due  to  its  momentum,  which  is  added  to  the  weight  pressure  of  the  water 
at  rest. 

3.  A  gram  mass  is  a  certain  fixed  quantity  of  material  and  a  gram 
weight  is  the  attraction  of  the  earth  for  this  mass.    This  attraction  changes 
with  certain  changes  in  location,  hence  the  weight  called  a  gram  weight 
is  not  always  the  same.    The  earth  is  not  a  true  sphere  and  it  is  rotating; 
these  are  the  conditions  which  lead  to  a  change  in  weight  with  changes 
in  location.    See  page  155. 

4.  A  spring  balance  measures  force  directly,  in  this  case  weight.     If 
the  spring  balance  is  being  used  in  the  same  latitude  in  which  it  is  gradu- 
ated the  mass  is  also  80  gm.  mass,  since  the   earth's  attraction  varies 
for  a  fixed  mass.     No. 

5.  Both  are  equal,  buoyancy  of  air  neglected.    Yes,  because  if  a  change 
in  location  changed  the  weight  of  one  it  would  produce  an  equal  change 
in  the  weight  of  the  other. 

6.  We  infer  that  something  else  is  acting  as  well  as  the  body  known. 
In  common  speech,  the  second  body  is  not  free  to  move. 

7.  Let /=  the  pull  of  the  engine. 


MOTION  15 


Let  a=the  acceleration  of  the  first  train. 
Then  ^  =  the  acceleration  of  the  second  train. 

Let  ra=the  mass  of  the  first  train. 
Let  m1=the  mass  of  the  second  train. 
Then  /=  ma. 


m=  2~* 
That  is,  the  mass  of  the  first  train  is  half  the  mass  of  the  second. 

8.  Let/  =  the  pull  of  the  earth  on  a  1-lb.  mass. 
Let  2/  =  the  pull  of  the  earth  on  a  2-lb.  mass. 
Let  a  =  the  acceleration  of  a  1-lb.  mass. 
Let  ac=the  acceleration  of  a  2-lb.  mass. 

Let  m=the  acceleration  of  a  1-lb.  mass. 
Let  2m  =  the  acceleration  of  a  2-lb.  mass. 
Then  2f=ma. 

or  2/=  2  ma. 
and  2/=2  mac. 
2  ma=2  mac. 

a=ac. 
That  is,  the  accelerations  are  equal. 

9.  1  gm.  weight  —  i.  e.,  the  weight  of  1  gm.  mass.     It  has   different 
values.    See  question  3.     1  dyne  (see  line  18,  page  98)  has  same  value 
everywhere  because   the   other  units  (cm.,  gm.  mass,  sec.)  used    in   its 
definition,  have  the  same  value  everywhere. 

10.  1  gm.  wt.  =  980.2  dynes.  7  gm.  =  6861.4  dynes.  7  gm.  wt.  in 
Panama  are  equivalent  to  a  less  number  of  dynes  because  the  earth  at- 
tracts a  gram  mass  less  there  than  it  does  at  New  York. 

MOTION—  Pages  109,  110 
1.  See  Fig.  3. 


2.  See  Fig.  4. 


16z=6  (44-x). 
22x=246. 


3=12. 

4ft 


44— a;- 32.  12  Ib. 

3.   When  A  is  on  pan  x 
and  B  is  on  y  (Fig.  5) . 

Am=Bn  (1). 

When  A  is  on  y  and  5  is  on  x 
An>Bm   (2). 


4ft. 


24  Ib. 


16 


MANUAL  FOR  MUMPER'S  PHYSICS 


Multiplying   (2)  by  (1) 

A2mn>&mn  (3). 
Dividing  (3)  by  mn 

AI>&  (4). 

A>B  (5). 
Dividing  (2)  by  (1) 

An^Bm 


FIG.  4 

n.   m 
orm>n 

Multiplying  by  mn,  n2>w2  (7) 
or  w>/w. 

4.  Horse  A  must  have  a  force 
arm  \  as  long  as  that  of  horse  B. 
Horse  A  must  have  J  of  the  en- 
tire length  and  horse  B  f  of  it. 
In  the  first  case  A  is  16  in.,  B, 
32   in.    from    point   of    attach- 
ment.   When  A's   force  is  f  of 
B's,  then  B  has  5  units  of  force  to 
A's  3,  and  A  has  f  of  the  dist., 
while  B  has  $  of  dist.    A's  dist. 
=  30  in.,  B's  =18  in. 

5.  10  lb. =  160  oz.    Let  x=dist.  of  sliding  wt.  from  support. 
160Xi=8:r. 

8z=80.  151b.  =  240oz. 


8z=120. 


FIG.  5 


6.  At  the  start  he  lifts  $  the  total  weight=i  of  250=125  Ib.     The 
rotational  moment  is  125X6=750,  the  ground  end  being  the  axis.     To 
rotate  it  around  same  axis  he  must  produce  the  same  moment  no  matter 
where  he  takes  hold.     If  the  arm  is  5  and  the  moment  750  the  force  = 
750  -^-5=  150  Ib. 

7.  See  Fig.  6. 

Let  a;=dist.  of  force  24  to  point  of  application. 
Then  12—  x=  dist.  of  force  18  to  point  of  application. 


12-X 


-> 


42 


24 


FIG.  6 


42x=2l6. 

v  The  resultant   has  a 

magnitude  of  42  gm.  and 
its  point  of  application 
is  5f  cm.  from  that  of 
24  gm. 


MOTION 


17 


36 


8.  See  Fig.  7. 

84  gm.— 36  gm.  =  48  gm., 
second  force. 

36X20=48*.    2t 

48x=720.  I  84 

x=l5. 

9.  See  Fig.  8.  * 
Each  boy   has  15  ft>. 

when  the  bucket  is  3  ft.  JTIG   7 

from  each  boy's  hand. 

When  x  has  20  Ib.  y  has  10  Ib.    Then  the  dist.  of  x  is  \  as  much  as 
that  of  y  from  the  bucket,  i.  e.,  x  has  2  ft.  and  y  4  ft. 

Whenx  has  24  Ib.  y  has  6 
J?     Ib.    Then  the  load  at  x  is  4 
ffft.  times  the  load  at  y,  hence  y's 

dist.  is  4  times  x's,  that  is,  x 
has  |  and  y  f  of  the  whole. 
|  of  6ft.=  l£=dist.  ofx.  And 
4X  H  =  4|ft.  =  dist.  of  y  from 
bucket. 

10.  There  is  an  increased 
tension  on  the  other  rope  of 
160-48=112  Ib.  48 -f- 160 X 
18=5.4  ft.,  dist.  from  man  to  the  rope  having  112-lb.  pull.  18—5.4= 
12.  6  ft.  dist.  from  man  to  the  rope  having  48-lb.  pull. 

MOTION— Page  113 

1.  Relative  to  the  earth  its  velocity  is  12 — 5=7  mi.  per  hr.  up  stream. 
ca  is    path  of   boat   relative 
to  bed  of  the    stream  and   cb  is 
the  path  through  the  water. 

Let  x=no.  hrs.  required  to 
cross. 

Then  xX  12= dist.  cb  the  boat 
travels  through  the  water  and 
xX5=dist.  ab  the  water  moves 
in  same  time;  but  in  the  right 
triangle,  abc,  (c6)2=  (ca)2  +  (a&)2; 
substituting,  (12x)2=(2)2+(5x)2. 

144x2=25x2+4. 
119x2=4. 
4 

*2=TI9- 
2 


FIG.  8 


11  min.  nearly. 


MAN.  MUM.  PHYS.— 2 


18 


MANUAL  FOR  MUMPER'S  PHYSICS 


2.  See  Fig.  10. 


&  =  (30)2  +  (40)2. 

#2  =900  +  1600  =2500. 

#=50  Ib. 


30  76. 


3.  See  Fig.  11.  #2=  (20)2  +  (4)2=  416. 

R  =  \/41G> 
#=20.4—  ft.  per  sec. 

4.  See  Fig.  12.  #2=  (25)2  +  (35)2=  1850. 

#2=V1850=43.0+   ib. 

5.  Since  the  direction  is  N.  W.  the  two 
components  are  equal.     Then  the  square 

of  the  northwesterly  speed  is  equal  to  twice  the  square  of  the  northerly 
speed. 

(16)2 


40Ib. 


FIG.  10 


4ft 


=  128. 

JV=<v/128=H-3+mi.  per  hr. 
The  westerly  speed  is  the  same. 
25lb.  6.  See  Fig.  13. 


2Oft. 
FIG.  11 


35/b. 


Since  the  angle  between  n  and  m  is  90°, 
and  n=m, 

2m?  =  (60)2 
3600 


FIG.  12 

7.  An    increase   in    the 

would  make  the  two  parts  more  nearly  parallel, 
hence  the  tension  would  become  more  nearly  £ 
the  wt.  of  the  picture.  A  decrease  makes  the 
angle  at  the  nail  greater  and  the  tension  on 
the  cord  must  become  greater  in  order  to  pro- 
duce the  required  resultant  which  must  always  be 
equal  to  the  weight  of  the  picture. 


m=42.4+lb.    on   each   part    of 

cord, 
length    of   the   cord 

<V 


FIG.  13 


MOTION— Page  119 

1.  By  balancing  the  bat  on  a  wire,  one's  finger,  etc.,  the  center  of 
gravity  may  be  found  roughly.     Or  it  may  be  suspended  by  means  of  a 
string  from  different  points.     A  certain  weight  of  wood  can  be  removed 
from  one  part  and  an  equal  weight  of  lead  or  iron  added  at  another  place. 

2.  It  may  be  balanced  as  suggested  in  Figs.  14  and  15.    If  he  measures 


ENERGY  AND  WORK 


19 


FlQ   14 


from  center  of  gravity  to  A  and  from  A  to  N  then  (wt.  of  hammer)  Xdist. 
from  A  to  cg=wt.  of  nailsXA/V.  From  this  the  weight  of  nails  can  be 
found. 

3.  On  account  of  its  great  density 
and   its   location  the    keel   lowers  the 
center  of  gravity  of   the  boat  without 
producing   much  effect   on  the   center 
of  buoyancy;  hence  it  makes  the  boat 
very  stable   and   enables   it  to  with- 
stand a  heavy  wind  without  upsetting. 

4.  A  boat  load  of   lumber  of  ne- 
cessity has  a  center  of  gravity  which  is  high,  nearly  as  high  as  the  center 

of  buoyancy,  hence  the  boat  is 
more  easily  rocked  and  upset  by 
the  winds. 

5.  The  lead  or  mercury  results 
in  the  center  of  gravity  being  near 
one  end,  the  lower,  so  that  it  will 
stand  erect  in  the  liquid.  Center 
of  gravity  is  below  the  center  of 

r  IG.   15 

buoyancy. 

6.  B,  on  account  of  the  lead  within  at  a,  has  its  center  of  gravity  very 
near  a,  hence  when  put  into  the  position  shown  in  the  cut  it  will  return 
to  the  erect  position  and  the  oil  cannot  spill.  But  in  can  A  the  center  of 
gravity  must  be  raised  in  order  to  place  the  can  upright. 


ENERGY  AND  WORK— Pages  127,  128 

1.  WoTk=FXD.    Work=3X2=6  ft.  Ib. 
3  ft.  =  3X30.48  cm.  =  91. 44  cm. 
91.44X200=18288.  gm.  cm.  of  work. 

2.  On  the  way  up  potential  energy  is  being  produced  from  kinetic. 
While  on  the  table  the  energy  of  the  book  is  said  to  be  potential  because 
it  is  not  energy  of  motion.    Our  right  to  say  that  it  has  energy  rests  upon 
the  fact  that  it  will  fall  if  the  supporting  table  is  withdrawn. 

3.  The  driving  of  a  nail  is  doing  work.    To  find  the  amount  of  work 
done  by  each  blow  we  must  know  the  average  force  with  which  the  ham- 
mer acts  and  the  distance  through  which  the  nail  is  moved. 

4.  More  when  striking  downward  for  in  this  direction  the  total  kinetic 
energy  is  the  sum  of  that  given  directly  by  the  workman  and  that  result- 
ing from  the  potential  energy  of  the  raised  hammer. 

6.  D  =  24   ft.     F=U()  ft).       In   lifting  his  body,   work=  24X140= 

3360  ft.  ft>.    When  lifting  the  box,  work  =24X25  =600  ft.  Ib.  additional. 

6.  Z>=7ft.     Force=  100+50=  150  Ib.     Work=  7X150=  1050  ft.   lb. 


20  MANUAL  FOR  MUMPER'S  PHYSICS 

The  force  the  boy  must  exert  in  pulling  the  wagon  containing  the  trunk 
is  much  less  than  the  combined  weight  of  trunk  and  wagon,  hence  he  does 
less  work  per  foot. 

The  force  the  boy  exerts,  measured  by  a  spring  balance,  and  the  dis- 
tance from  one  street  corner  to  the  next  must  be  known. 

7.  Three  cu.  ft.  of  water  weigh  3X62.4  lb.=  187.2  Ib.     1  yd.  =  3  ft. 
Work=  3X187.2  =  561. 6  ft.  Ib.     The  amount  of  work  required  to  lift  the 
water  a  given  distance  is  not  affected  by  the  direction  provided  no  other 
factors  are  introduced  thereby  and  the  vertical  elevation  is  the  same. 

8.  (When  is  work  or  energy  said  to  be  transferred?)     When  energy  is 
given  from  one  body  to  another.    It  is  transformed  when  it  is  changed  from 
one  kind  into  another,  as  for  example,  mechanical  energy  into  heat. 

9.  Distance  lifted  is  1  ft.  in  12,  hence  5  ft.  in  60  ft.    Work=  5X 1000  = 
5000  ft.  Ib.  (not  counting  waste).     The  arrangement  of  the  trunks  af- 
fects the  amount  the  man  must  lift  but  not  the  force  with  which  he  pushes, 
hence  the  amount  of  work  is  not  affected. 

10.  (a)  Work=  3X400  =1200  ft.  Ib. 
(6)  Work=3X650=1950  ft.  Ib. 

11.  The  energy  of  his  moving  hand  is  called  kinetic;  that  of  the  wound 
spring  is  called  potential.    It  is  called  potential  because  it  is  not  energy 
of  motion  but  may  be  converted  in  energy  of  that  type. 

12.  We  must  know  the  force  with  which  the  powder  acts  and  the  dis- 
tance through  which  it  acts  upon  the  bullet.     (Approximately  the  length 
of  the  barrel.)    The  weight  of  the  bullet  and  the  distance  it  rises  before 
it  comes  to  a  stop.    At  the  instant  it  stops,  in  the  air,  its  energy  is  called 
potential.    At  the  instant  of  leaving  the  gun  its  energy  is  kinetic  and  it 
may  be  computed  by  knowing  the  mass  and  the  velocity.     E=\  MV2. 


13.  E=^p(pagel26).     £=g^=20  ft.  Ib. 

250X(60)2 
E==  2(980-}-)  =459'18  gm-  cm- 

14.  The  most  kinetic  energy  when  it  is  at  the  lowest  point  where  its 
speed  is  greatest  and  the  most  potential  at  the  highest  point  of  the  arc 
where  it  comes  to  rest. 

When  at  rest  at  its  lowest  point  it  has  no  energy  as  a  pendulum,  though 
it  is  said  to  have  potential  energy  as  a  body  elevated  above  the  surface 
of  the  earth. 

16.  Kinetic  energy=£  MV2.  Since  motion  and  speed  are  relative  to 
some  body;  hence  energy  as  thus  computed,  depends  upon  the  standard  of 
motion.  Potential  energy  computed  from  position  must  be  relative  for 
position  is  relative.  See  pages  88-89. 

16.  We  usually  select  a  point  on  the  earth's  surface  vertically  below 
the  object. 

17.  Work = energy  rendered  potential =800  XI 500  =1200000  gm.  cm. 


ENERGY  AND  WORK  21 

The  potential  energy  will  have  become  kinetic,  hence  the  body  when  strik- 
ing has  1200000  gm.  cm.  of  energy. 

d=' 


v=gt    (page  93) 


H 


=  V2X  1.500X980 

=  \/2940000 

=  1714.6  cm.  per  sec. 

18.  A  pound  force  changes  slightly  with  a  change  in  latitude,  but  a 
dyne  has  the  same  value  everywhere.  In  consequence  a  foot  pound 
varies  with  latitude,  but  an  erg  has  the  same  value  everywhere.  The 
latter  is  based  on  the  gm.  (mass),  cm.  (length),  and  second  (time). 

ENERGY  AND  WORK— Pages  131,  132 

1.  Work  to  be  done = force  Xdist. 
Force  is  1000X8  lb.  =  8000  lb. 
Work=  8000X20=  160000  ft.  lb. 

The  amount  of  work  is  the  same  in  each  case. 
A  does  it  in  40  min.  or  1600004-40=4000  ft.  lb.  per  min. 
B  does  it  in  60  min.  or  1 60000 -h  60  =2666f  ft.  lb.  per  min. 
B's  power  is  only  §  that  of  A. 

2.  A  one-horse  power  engine  can  do  33000  ft.  lb.  per  min.,  hence  a 

4  h.  p.  engine  can  do  4X33000=  132000  ft.  lb.  per  min. 
3  hours=  3X60  =180  min. 
A  4  h.  p.  engine  can  do  180X132000=23760000  ft.  lb.  in  3  hours. 

3.  The  work  to  be  done  is  160000  ft.  lb. 

A  2  h.  p.  engine  can  do  66000  ft.  lb.  in  1.  min. 

To  do  160000  ft.  lb.  it  will  require  160000 -H 66000=  2.4+  min. 

4.  10000  cu.  ft.  of  water  weigh  624000  lb. 

Work  to  be  done  in  1  hour  =60X624000  =37440000  ft.  lb. 

Work  to  be  done  in  1  min.=  5V  of  37440000=624000  ft.  lb. 

1  h.  p.  can  do  33000  in  1  min.,  hence  to  do  624000  ft.  lb.  in  1  min. 

will  require  624000-1-33000=18.9  h.  p. 
6.   1  gm.  wt.  =  980.+  dynes.    To  lift  1  gm.  4  cm.  requires  4X980= 

3920  ergs.     See  page  129. 
6.   1  kilowatt=1000  watts  =  1000  joules  per  sec. 

10  kilowatts=  10000  joules  per  sec. 

15  min.  =  900  sec.     10  kilowatts  for  15  min.  =  9000000  joules. 

1  kilowatt^  H  h.  p.     10  kilowatts^  13$  h.  p. 


22  MANUAL  FOR  MUMPER'S  PHYSICS 

7.  The  wasted  work  appears  chiefly  as  heat  at  the  axles  and  platform 
on  which  the  truck  runs. 

Oiling  the  axle  decreases  the  waste  work  but  does  not  affect  the  useful 
work.  Ball  bearings  decrease  the  waste  work  or  heat  at  the  axles,  hence 
decreases  the  total  amount  of  work  the  man  must  do. 

8.  When  roads  are  graded  and  smooth  a  given  power  can  do  more 
useful  work,  hence  a  horse,  for  example,  can  move  more  material  with 
the  same  expenditure  of  energy.    The  saving  in  power  usually  more  than 
makes  up  for  the  cost  of  improvement. 

9.  To  do  work  upon  itself  as  a  whole,  a  body  would  have  to  create 
energy. 

One  part  may  do  work  upon  another  part  of  the  same  body.  The  loss 
of  energy  in  one  part  equals  the  gain  in  the  other.  A  man  may  use  one 
hand  to  lift  the  other. 

ENERGY  AND  WORK— Page  136 

1.  It  is  no  evidence  of  another  body  acting  on  the  earth.    (First  law.) 
It  shows  attraction  between  the  parts  of  the  earth  (cohesion)  or  each 
particle  would  move  in  a  straight  line  tangential  to  the  circumference. 

2.  Yes,  otherwise  the  earth  would  move  in  a  straight  line. 

3.  To  do  so  would  be  an  object  putting  itself  in  motion  or  creating 
its  own  energy.     (First  law.) 

4.  When  the  shoe  stops,  the  adhesion  of  the  snow  to  the  shoe  is  not 
strong  enough  to  overcome  the  inertia  of  the  snow,  which  therefore  moves 
on  away  from  the  shoe. 

MACHINES— Pages  139,  140 

1.  It  is  said  to  furnish  a  gain  in  force  when  by  its  use  the  force  upon 
a  load  is  made  greater  than  it  is  at  the  agent.    It  furnishes  a  gain  in  speed 
when  the  speed  of  the  load  is  greater  than  that  of  the  agent.    A  change 
in  direction  means  that  the  load  moves  in  a  different  direction  from  that 
of  the  agent.    A  gain  in  force  is  accompanied  by  a  loss  in  distance  moved 
by  the  load,  that  is,  loss  in  speed  of  load.    See  sec.  85,  page  139. 

2.  Simple   machines   are   intended    only   to   transmit   energy.      The 
energy  they  transform  is  wasted,  generally  as  heat. 

3.  It  can  give  out  only  the  energy  given  to  it,  and  since  it  gives  as 
fast  as  it  receives,  it  is  in  no  sense  a  storehouse  of  energy. 

4.  Since  they  transmit  energy  and  nothing  more  they  cannot  add 
to  the  quantity  given  them  per  unit  of  time  (by  any  agent  as  a  man  or 
engine).    They  may  multiply  force  or  speed.    See  ans.  to  ques.  1. 

5.  See  3,  page  138. 

Oiling  decreases  the  amount  of  the  waste  energy  and  also  diminishes 
the  wear  on  the  bearings. 


MACHINES  23 


MACHINES— Pages  144,  145,  146 

1.  See  Fig.  16.     AB  =  28  in.     AC=U  in.     DC  =7  in. 

BC=U  in. 

ACXW=BCXx.  A  C  D          B 

14X40=  14Xz.  i  A  '  I 

The  agent  at  B  must  act  down- 
ward with  a  force  of  40  Ib. 

2d  case.     AC X 40= DC X?/. 
14X40=7X2/. 
7j/=560. 
7/=80. 
An  agent  at  D  must  act  downward  with  a  force  of  80  Ib. 

2.  First  class.    Place  the  load  at  D  acting  downward  and  agent  at  B 
acting  upward,  and  it  shows  second  class.    Place  load  at  B  acting  upward 
and  agent  at  D  acting  downward,  and  it  shows  third  class. 

3.  See  page  139,  first  5  lines. 

4.  The  agent  must  do  an  amount  of  work  (varying  with  circum- 
stances) more  than^ijiat  computed  as  necessary  to  move  the  load. 

6.  Since  AO,  agent  arm,  is  shorter  than  BO,  load  arm,  the  force  at 
A  must  be  greater  than  that  produced  at  B,  hence  there  is  a  loss  in  force. 
But  the  load  moves  a  greater  distance  in  same  time,  hence  it  has  a  greater 
speed  than  that  of  agent. 

6.  The  load  is  2  ft.  from  axis,  the  agent  is  2+3=5  ft.  from  axis. 
2X225=  5Xz.     5z=450. 

z=90  Ib.  at  man's  hands. 

7.  It  will  increase  the  amount  the  man  must  lift.     160X1  =  5Xy. 
5^=160.      ?/=32  Ib.,  due  to  wt.  of  wheelbarrow.     90  Ib.  (previous  prob- 
lem)+32=  122  lb.  =  total  force  at  man's  hands,  61  Ib,  on  each  hand. 

8.  (a)  At  the  axle,  225-90=135  Ib. 

(6)  Due  to  wt.  of  barrow,  160  -32=  128  Ib. 

Due  to  both  load  and  barrow,  135  lb.+128  lb.  =  263  Ib. 

9.  Sugar  tongs,  agent  at  T,  fulcrum  at  B,  load  at  S. 
Scissors,  agent  at  F,  fulcrum  at  A,  load  at  C. 

Claw  hammer,  agent  at  end  of  handle,  fulcrum  where  hammer 
head  touches  the  wood,  load  at  nail. 

10.  Agent  at  keys,  load  the  type. 

Nut  cracker,  agent  at  C,  fulcrum  at  A,  load  at  B. 

11.  Not  counting  waste,  amounts  are  equal.     See  page   139.     When 
grasped  nearer  the  head  the  agent  must  act  with  more  force  to  produce 
the  required  force  at  the  nail  or  load.    The  work  is  the  same,  a  change 
in  direction  of  action  would  change  force  but  not  the  useful  work  required, 
though  on  account  of  increased  friction  it  would  increase  the  wasted  work. 


24 


MANUAL  FOR  MUMPER'S  PHYSICS 


30/b, 


12.  See  Fig.  17.  FB=S 
ft.      FA  is  to  be  found. 
FAX  5Q=FBX30. 


FIG 


ft.    To  find  FA  (not  counting  the  wt.  of  plank); 
FA  is  f  and  BF  f  of  the     A  ^ 

whole,  7^4  =  6  ft. 

To  find  FA  (counting 
the  wt.  of  plank)  .  Center 
of  gravity  (C)  is  in  mid- 
dle of  plank. 

See  Fig.  19. 


FA  =  4.  8  ft. 

See   Fig.    18.     AB=16 
=  3QXBF,  hence 


50/b. 


3Olb. 


FIG.  18 


A                          F    C 

B 

50/b.              A 

\30Ib. 

Let  x  =  dist.  from  C  to  F. 
S—x  =  dist.  from  A  to  F. 
=  dist.  from  B  to  F. 


FIG.  19 


50  X^ 

CF. 

50  (8  —  x)  =  30  (8  +x)  +40z. 
120z=160. 


13.  The  arm  is  a  lever  having  its  axis  at  y  and  agent,  the  muscle,  acting 
at  x.  The  force  on  the  load  w  is  much  less  than  the  force  or  tension  of 
the  muscle  at  x,  but  there  is  a  corresponding  gain  in  the  speed  of  the 
load,  a  very  important  advantage.  Not  counting  the  waste  the  work 
done  by  the  muscle  is  equal  to  that  done  upon  the  load  W. 


FIG.  20 


MACHINES— Page  148 

1.  The  speed  of  the  load 
is  here  equal   to   that   of   the 
agent.     It  furnishes  a  change 
in  the  direction  of  motion  of 
the  load  in  relation  to  that  of 
the  agent. 

2.  See  Fig.  20.    The  force 
at  the  load  is   greater.     The 
distance    the     agent     moves 
is    greater.      FaXDa=FlXDl 
(waste  neglected). 

3.  See  Figs.  21  and  22. 

4.  In  the  first  case  (Fig.  21) 
force  A :  force  L=  1 :  2.    In  the 


FIG.  21 


MACHINES 


25 


FIG.  22 


FIG.  23 


second  case  (Fig.  22)   force 

A:  force  L=l:  3.       In  first 

case    the    force  at    load   is 

2X15  ft.  =301b.andwork= 

8X30=  240  ft.  lb.     In  second 

case  force  at   load=3Xl5= 

45  lb.  and  work=  8X45=  360 

ft.    ft>.     These    results    give 

the  work  upon  the  load  alone. 

The  agent  must  do   an  ad- 
ditional amount  to  make  up 

for  waste. 

5.  See  Fig.  23.    Force  at 

L  is  5  times  the  force  at  A, 

hence  force   L=  5X210  gm. 

=  1050  gm.  If  fixed  end  of 
cord  is  attached  to  fixed  block  of  pulleys  there  will  then  be  only  4  parts 
of  cord  to  movable  block  (invert  the  figure  shown),  hence  force  at  L= 
4X210  =  840  gm.  (not  counting  weight  of  pulleys). 

MACHINES— Page  150 

1.  The  radii  are  10  cm.  and  3  cm. 
Force  at  AX  10  =  force  at  LX3. 

LX3=300X10. 
3L  =  3000. 
L=1000  gm. 

2.  The  distance  the  agent  moves,  hence  the  speed  of  the  agent,  is 
3£  times  as  great  as  that  of  the  load.     Work  on  L=  15X1000=15,000 
gm.  cm.;  neglecting  waste,  the  quantities  of  work  are  equal  according  to 
the  law  of  conservation  of  energy. 

3.  96X^=4.5X480 

4.5X480 

96 
=  22.5  in.,  length  of  crank. 

4.  The  large  wheel  has  5  times  the  diam.  and  5  times  the  circum- 
ference of  the  small  wheel.    The  chain  moves  enough  in  1  rotation  of  the 
large  wheel  to  rotate  the  small  wheel  5  times. 

5.  There  is  a  gain  in  speed  with  an  accompanying  loss  in  force. 

MACHINES— Pages  153,  154 

1.  The  weight  of  the  wagon  is  lifted  through  the  vertical  height  of 
the  hill,  but  the  horse  exerts  his  force  throughout  the  entire  length  of  the 
hill. 


26  MANUAL  FOR  MUMPER'S  PHYSICS 

2.  The  force  with  which  a  body  acts  in  a  given  direction  and  the 
distance  through  which  it  moves  another  body  in  that  direction.    When  a 
body  is  being  lifted  the  weight  is  the  required  force  and  the  vertical  dis- 
tance is  the  required  distance. 

3.  We  must  know  the  pull  (in  Ibs.,  dynes,  etc.)  exerted  by  the  horse 
and  the  distance  through  which  the  horse  moves  the  wagon  measured  on  the 
slope  of  the  hill.     Or  if  we  know  the  weight  of  the  wagon  and  the  vertical 
distance  we  may  compute  the  work   from  them.     Neglecting    waste  the 
results  should  be  equal. 

4.  Distance  lifted,  20  ft.Xwt.  of  wagon,  500  lb.=  10000  ft.  lb.  =  work. 

5.  The    horse  must  do  10000  ft.  Ib.  in  moving  the  wagon  100  ft., 
hence  his  force  is  10000  -M00=  100  Ib. 

6.  Here  " easier"  means  less  force  required.     Work  is  the  same. 

7.  In  this  case  "  easier  "  means  less  work,  for  the  oiling  decreases  the 
waste,  hence  the  total  work  the  horse  must  do  is  decreased. 

8.  (a)   250  Ib.  (6)  T57  of  250=  83£  Ib. 

(c)   Work,  when  moved  directly  =250X5=  1250  ft.  Ib. 

Work,  when  moved  on  the  plank  =83^X15  =1250  ft.  Ib. 

9.  Area  of  small  piston,  20  sq.  cm. 
Area  of  large  piston,  180  sq.  cm. 

Total  pressure  on  small  piston  =60  gm.  . 

Total  pressure  on  large  piston  =540  gm. 

Work  at  small,  piston =45X60  =2700  gm.  cm. 

Work  at  large  piston  is  also  2700  gm.  cm.  (waste  neglected). 

Distance  large  piston  moves  =2700-^540  =5  cm. 

MOTION  IN  A  CURVED  LINE— Page  158 

1.  When  the  speed  reaches  a  certain  amount  the  cohesion  of  the 
particles  of  the  wheel  is  no  longer  able  to  pull  the  parts  of  the  wheel  into 
curved  paths,  hence  they  move  off  in  (approximately)  straight  lines. 

2.  When  changing  direction  there  must  be  an  action  and  reaction 
between  the  wheels  and  the  roadbed;    if  on  account  of  mud,  etc.,  this 
is  insufficient  at  any  part  of  the  curve  the  machine,  or  a  part  of  it,  may 
suddenly  resume  the  straight-line  motion. 

3.  It  makes  the  weight  of  a  body  less  or  apparently  less  than  it  other- 
wise would  be.    It  is  greatest  at  the  equator  and  zero  at  the  poles. 

,    mi  in   (massX  veloc.2)  . 

4.  The  upward  pull   *  — — Q -  is  equal  to  or  greater  than  the 

downward  pull  (weight  of  the  mass). 

5.  The  same  as  4. 

6.  It  has  a  large  amount  of  kinetic  energy.     This  stock  of  energy 
may  be  drawn  upon  or  added  to  in  the  working  of  the  machine  without 
materially  affecting  the  speed,  because  the  mass  is  so  large. 


HEAT  27 

THE  PENDULUM—  Pages  161,  162 

1.  The  earth  pull  or  the  weight  of  the  bob.    It  vibrates  in  less  time 
when  near  the  poles  because  the  value  of  g  is  greater. 

2.  The  mainspring  is  the  storehouse  of  the  energy  given  to  the  clock 
when  it  is  being  wound;  it  produces  the  motion  of  the  wheel  work.    The 
pendulum  regulates  the  motion,  that  is,  it  prevents  the  wheels  from  going 
too  fast.     If  it  runs  too  fast  make  the  pendulum  longer,  so  that  more 
time  is  used  in  making  each  vibration. 

3.  *2=f  sec.  ti:t2=VTi-  Vh. 


squaring  1  :  f  =  39.  1  :  £2. 
Z2=17.3+in. 

4.  tj?:W=li:h. 
(3)2;  (4)2=25:  h 
9Z2=  400. 

Z2=44.4  cm. 

5.  \~  =  Y-  or  ^  =  ^  ,  substituting 

*         ^92      0i    gz 

™.       99.3^=98020. 
<72=  987.1. 

6.  W:t#=h:l2. 
1:  16=36:  Z2. 

12=  16X36=  576  in. 

7.  An  increase  in  earth's  attraction  or  in  the  value  of  g  would  decrease 
the  time  required    by  a  given  pendulum  to    make  one  vibration.     This 
suggests  that  a  given  pendulum,  of   fixed  length,  may  be  used  to  de- 
termine the  relative  values  of  g  at  different  places. 

HEAT—  Page  175 

1.  4°  C.  are  equivalent  to  4Xf°  F.  =  7i°  F. 

2.  The  body  is  above  zero  and  it  is  warmer  than  freezing  water. 

3.  It  is  above  zero  F.,  but  it  is  28  F.  degrees  below  the  temperature 
of  freezing  water. 

4.  See  sec.  105,  page  170. 

5.  68°  F.—  32°  F.  =  36  F.°.    This  means-  that  the  room  is  36  F.  degrees 
above  the  freezing  point.     36°  F.=  f  of  36=  20°  C. 

6.  f  of  45°  =  81°  F.    This  is  the  equivalent  number  of  degrees.     But 
the  reading  will  be  81°+32=113°  F.  on  the  scale. 

7.  4°C.  =  fX4+32=39°-f  F. 
10°  C.  =  |X10+32=50°  F. 

-40°  C.=  »X  -40+32=  -40°  F. 


28  MANUAL  FOR  MUMPER'S  PHYSICS 

98°  F.=  f(98-32)  =  36.°+  C. 
10°  F.  =  f(10— 32)=— 12.°  +  C. 
-10°  F.=  f(-10-32)=-23.°  +  C. 

8.  —  273°  C.=  |X  -273+32=  —459.°+  F. 

9.  -182°  C.=  |X  -182+32=  -295°+  F. 

10.  The  bulb  must  be  relatively  large  and  the  bore  of  the  stem  relatively 
small.  A  large  bulb  heats  and  cools  slowly.  The  instrument  is  not  re- 
sponsive. When  the  bore  of  the  tube  is  very  small  it  is  difficult  to  see 
the  mercury. 

HEAT— Pages  179,  180 

1.  10X1  gm.  cal.  =  10  gm.  cal.     10X25  gm.  cal.  =  250  gm.  cal. 
17°— 11°  =  6°  rise  in  temperature.     6X65  gm.  cal.  =  390  gm.  cal. 
95°— 15°=  80°  fall  in  temperature.    30X 80  gm.  cal.  =  2400  gm.  cal. 

2.  The  quantity  of  heat  lost  by  the  warmer  equals  that  gained  by 
the  colder. 

3.  (a)  The  temperature  is  the  mean  of  the  two  others;  that  is,  it  lies 
halfway  between  them. 

(6)  The  temperature  of  the  colder  and  smaller  mass  rises  twice  as 
many  degrees  as  that  of  larger  mass.  Hence  the  rise  is  §  and  the  fall 
^  of  "the  difference  between  the  original  temperatures. 

(c)  The  smaller  changes  f  and  the  larger  f  of  the  number  of  degrees 
difference  in  their  original  temperatures. 

4.  Because,   being  different  substances  they  have  different  specific 
heats  and  when  one  gains  as  much  heat  as  the  other  loses  they  do  not 
undergo  equal  temperature  changes. 

5.  The  mass,  the  specific  heat  and  the  temperature  change. 

6.  The  rise  in  temperature  equals  the  fall  in  temperature,  hence  they 
come  to  a  final  temperature  halfway  between  the  original  temperatures — 
£(0+56)  =  28°  C. 

7.  87°— 23°  =  64°  C.,  temperature  difference.     The  rise  in  tempera- 
ture equals  the  fall  in  temperature,  and  the  final  is  23° +32°=  55°. 

8.  Temperature  changes  are  inversely  as  the  masses.     58° — 14°  =44°, 
whole  difference  in  temperature. 

Rise  in  temperature  is  T8T  and  fall  is  T6¥  of  the  whole  difference,  44°. 

Rise=T8¥of  44=25f°. 

Final  temperature  =  14° +25  j°  =  39|°  C. 

9.  It  takes  .11  of  a  gm.  cal.  to  warm  1  gm.  of  iron  1°  C.,  or  more  gen- 
erally, it  takes  .11  times  as  much  heat  to  warm  any  mass  of  iron  1°  as 
it  does  to  warm  an  equal  mass  of  water  1°. 

10.  It  does  not,  because  specific  heat  is  a  ratio. 

11.  Heat  given=25Xl  gm.  cal.  =  25  gm.  cal. 

1  gm.  of  iron  requires  .11  gm.  cal.  for  each  degree. 


HEAT  29 

10  gm.  of  iron  require  1.1  gm.  cal.  for  each  degree. 
25-M. 1  =  22.7°  +  C. 

12.  80°— 15°  =  65°,  the  whole  temperature  difference. 
Let  3= fall  of  temperature  of  iron. 

65— 3  =  rise  in  temperature  of  water. 

Quantity  of  heat  lost  by  iron  =  quantity  gained  by  water. 

.11X3203=320(65—3). 

.113=65-3. 

x  =  58.5°,  fall  in  temperature  of  iron, 
hence  final  temperature =80°—  58. 5°  =  21.5°. 

13.  75°— 15°  =  60°,  total  temperature  difference. 

Let  3= rise  in  temperature  of  water  and  of  glass  vessel. 
60— 3= fall  in  temperature  of  copper. 
Heat  received  by  water  and  vessel = heat  lost  by  copper. 
2003 +  .16(403)  =  .09X450(60 —3). 
2003+6.43  =  2430—40.53. 
246.93=2430. 

3=9.8°  C. 
final  temperature=150+9.8°  =  24.8°  C. 

14.  Let  3= specific  heat  of  metal. 

100°— 30°  =  fall  in  temperature  of  metal=70°. 
30°— 16°  =  rise  in  temperature  of  water  =14°. 
720X703=120X14. 
504003=1680. 

3=  .033  +  =  specific  heat  of  metal. 

16.  It  would  warm  1  gm.  of  water  through  .091°.    The  same  amount 
of  heat  would  warm  .091  gm.  of  water  through  1°. 

16.  To  warm   100  gm.  of  brass   1°  C.  require   100X.091  =  9.1  gm.  cal. 
The  same  quantity  of  heat  would  warm  9.1  gm.  of  water  through  1°.    The 
water  equivalent  of  100  gm.  of  brass  is  9.1  gm.  of  water. 

17.  200  gm.  of  lead  require  200X-031  =  6.2  gm.  cal.  to  warm  it  1°. 
6.2  gm.  of  water  would  require  the  same  amount  of  heat  to  warm  it  1°. 
Water  equivalent  of  200  gm.  of  lead  is  6.2  gm.  of  water. 


HEAT— Pages  185,  186 

1.  (a)  The  longer  increases  in  length  three  times  as  much  as  the 
other  increases. 

(6)  Because  they  are  in  contact  with  each  other.  It  could  be  deter- 
mined by  means  of  a  thermometer. 

(c)  Since  the  longer  increases  three  times  as  much  as  the  other,  the 
increase  of  each  is  the  same  part  of  the  original  length  of  each. 

(d)  Since  the  expansion  of  each  is  the  same  part  of  the  original  length 


30  MANUAL  FOR  MUMPER'S  PHYSICS 

of  each  and  they  are  warmed  through  the  same  number  of  degrees,  the 
increase  in  length  per  unit  length,  called  the  coefficient  of  expansion,  is 
the  same. 

(e)  This  expansion  would  be  larger  when  the  degree  is  a  centigrade 
degree,  than  it  would  be  if  it  were  a  Fahrenheit  degree. 

2.  Original  length,  original  temperature,  final  temperature  and  amount 
of  expansion  of  body. 

Yes,  because  if  the  longer  rail  were  used  both  the  original  length  and 
the  amount  of  expansion  would  be  three  times  as  large  and  their  quotient 
would  be  the  same  as  that  obtained  with  the  shorter  rail. 

3.  Any  piece  of  glass,  of  that  kind,  increases  or  decreases  in  length 
.0000089  of  its  original  length  for  each  degree  C.  change  in  tempera- 
ture. 

4.  (a)  200  X-  0000231  =.00462  cm.  for  1°  C. 
(6)  .00462X25=  .1155  cm.  for  25°  C. 

200  +  .  1155=  200.1155  cm.  length  at  25°  C. 

5.  Let  x  =  coefficient  of  expansion 
654.55—  654.00=.  55  cm.  increase  for  100° 

654XlOOz=.55 
65400*  =.55 


65400 
=  .0000084. 

6.  Expansion  decreases  and  contraction  increases  the  density  of  a 
body. 

Between  0°  C.  and  4°  C.  water  expands  as  it  cools  and  contracts  as 
it  gets  warmer,  hence  warming  water  between  these  temperatures  in- 
creases and  cooling  decreases  its  density. 

7.  1  m.=  100  cm. 

100  X  120  X.  0000089  =.1068  cm. 

100+.1068  cm.=  100.1068  cm.,  length  at  120°  C. 

8.  From—  8°  to  +37°  C.  is  a  temperature  change  of  45°  C. 
180.2X45X-  0000168=.  136+  ft. 

180.2+.136  ft.  =  180.336  ft.,  length  at  37.°  C. 

00001  87 

9.  Brass  expands  :Qoool2T=  1-54  times  as  much  as  iron  per  unit  length. 

Iron  rod  must  be  1.54X26.8=41.27  cm.  long  to  equal  the  expansion 
of  brass  rod  26.8  cm.  long. 

10.  Temperature  change,  86°—  5°  =  81°  F.  or  f  of  81  =  45°  C. 
300  X  45  X.  0000109=.  147  ft.,  expansion  of  span. 

11.  .0000297—  .0000  168  =.0000  129,   hence  each  cm.   of  zinc  expands 
.0000129  cm.  more  for  1°  C. 

600  X  60  X.  0000  129  =.46+   cm,,  the  excess  expansion  of  600  cm.  of 
zinc  for  60°. 


HEAT  31 

—  ---- 

HEAT—  Page  192 

1.  The  coefficient  of  expansion  of  gases  is  273-  f°r  eacn  1°  C.     The 
expansion  in  this  case  is  1  c.  c.  and  the  volume  becomes  274  c.  c.    In  the 
second  case  the  expansion  is  ^W  of  273=40  c.  c.,  and  the  volume  becomes 
273  +40=  313  c.  c.  at  40°  C. 

2.  If  originally  at  0°  C.  it  must  be  warmed  273°  to  expand  f^f  or  to 
double  its  volume. 

In  cooling  to  —63°  it  contracts  $fo  of  its  volume,  hence  its  volume 
is  then  Iff—  ^  =  fM  of  its  magnitude  at  0°  C. 

3.  The  pressure  of  the  gas  increases.    See  189,  sec.  119,  and  last  4  lines 
on  page  190. 

4.  0°  C.  =  273  on  absolute  scale. 
273+91  =  364°  on  absolute  scale. 
Pressure  becomes  fff  of  original=l$. 

In  terms  of  barometer  height  the  pressure  is  1^X76=101.3  cm. 

6.  The  pressure  being  constant  the  volume  varies  directly  as  the 

absolute  temperature,  hence  at  20°  C.,  or  293°  absolute,  the  pressure  is 

If  |  of  the  original  pressure;  and  the  density,  which  varies  inversely  as  the 

volume,  is  f$f  of  1.293=  1.204  gm.  per  liter. 

6.  60°C.  =  333°  absolute. 
180°  C.  =  453°  absolute. 
453  :  333  =  300  :  x. 

453z=  99900. 

x=  220.5  c.  c.  vol.  at  60°  C. 

7.  See  page  64,  sec.  39. 


1X76     F2X70 

273  ~     293    ' 

V2=  1.16  liters. 

Since  1.293  gm.  now  have  a  vol.  of  1.16  liters  the  density  is  T 
1.114  gm.  per  liter. 
0    FiPi     F2P2 


500X75     80XF2 
283  293    * 

F2=485.+  c.  c. 

9.  He  inhales  a  greater  weight  of  air  when  the  pressure  is  high;  a 
greater  weight  when  the  temperature  is  low;  a  greater  weight  when  at 
the  sea  level;  a  greater  weight  in  winter,  other  conditions  being  the  same 
in  all  the  cases. 

HEAT— Pages  197,  198 

1.  The  water  loses  as  much  heat  as  the  ice  receives.  The  temperature 
of  the  water  falls,  but  that  of  the  ice  does  not  change  because  the  heat 


32  MANUAL  FOR  MUMPER'S  PHYSICS 

received  by  the  ice  does  the  work  of  melting  and  hence  ceases  to  be 
heat. 

2.  1  gm.  of  lead  at  its  melting  point  requires  5.9  gm.  cal.  to  melt  it. 
The  thermometer  makes  a  difference  because  a  heat  unit  as  defined  by 
means  of  a  C.  thermometer  does  not  have  the  same  value  as  a  unit  de- 
nned in  terms  of  the  F.  thermometer. 

3.  80  gm.  cal.  per  gm.  of  ice,  hence  10  gm.  of  ice  require  800  gm.  cal. 
10  gm.  of  lead  require  10X5.9=59.  gm.  cal. 

4.  It  is  losing  heat  without  becoming  colder.    The  molecular  motion 
of  the  water  furnishes  the  heat.    The  heat  which  leaves  the  water  is  taken 
in  by  the  surrounding  colder  air. 

5.  60°  —20°  =  40°.    In  cooling  40°  the  water  gives  out  40  X  250=  10000 
gm.  cal. 

1  gm.  of  ice  requires  80  gm.  cal.  to  melt  it. 

10000  gm.  cal.  will  melt  10000-1-80=125  gm.  of  ice. 

6.  120  gm.  cal.— 80  gm.  cal.  =  40  gm.  cal.,  the  amount  of  heat  left  to 
warm  the  water;  this  will  warm  1  gm.  of  water  40°  C. 

As  the  ice  melts  the  volume  decreases  and  as  the  water  is  warmed 
from  0°  it  decreases  until  a  temperature  of  4°  C.  is  reached.  It  will  ex- 
pand from  this  temperature  up. 

7.  The  ice  is  better  because  it  requires  a  large  amount  of  heat  to  melt 
it  in  addition  to  the  heat  required  to  warm  the  ice  water  afterward  up  to 
the  highest  temperature  permitted  in  the  refrigerator. 

8.  It  takes  4  times  as  much  heat  to  melt  a  gm.,  or  any  other  mass  of 
ice,  as  it  does  to  warm  an  equal  mass  of  water  20°,  hence  1  Ib.  when  melted 
and  warmed  20°  would  require  as  much  heat  as  5  Ib.  of  ice  water  require 
in  being  warmed  from  0  to  20°. 

9.  Heat  lost  by  water  and  calorimeter  =  heat  used  to  melt  ice  and 
warm  water. 

Let  z=fall  in  temperature  of  water  and  calorimeter. 
Then  200z +. 09  X40z= 60X80 +60X10. 

z=26.5°. 

10° +265°  =  36.5°,    original   temperature   of  the  water 
and  calorimeter. 

HEAT— Pages  208,  209 

1.  From  liquid  to  vapor  state.    It  is  receiving  heat,  but  it  is  not  getting 
warmer.    The  heat  received  is  used  in  doing  the  work  of  vaporization. 

2.  1  gm.  of  water,  536  gm.  cal. 
8  gm.  of  water,  4288  gm.  cal. 
1  gm.  of  ether,  90  gm.  cal. 

12  gm.  of  ether,   1080  gm.  cal. 

3.  1  gm.  of  steam  by  condensation  alone  furnishes  536  gm.  cal.    16  gm. 


HEAT  33 

of  steam  furnishes  16X536=8576  gm.  cal.    To  warm  400  gin.  of  water  1° 
requires  400  gm.  cal.,  hence  8576  gm.  cal.  will  warm  the  water  -^-=21.4°. 

The  temperature  of  the  water  must  have  been  100°— 21.4°  =  78.6°  C. 
before  the  steam  was  admitted. 

4.  The  water  is  not  warmed,  all  the  heat  given  out  by  the  iron  is  used 
in  vaporizing  water.    The  iron  cools  800° — 100°  =  700° .    It  gives  out  700  X 
1000 X.I  1  =  77000  gm.  cal.     This  heat  will  vaporize  77000-^-536=143.+ 
gm.  of  water. 

5.  (1)  100°  C.    (2)  40°  C.    (3)  The  heat  is  furnished  both  by  the  con- 
densation of  the  steam  and  the  cooling  of  the  water  formed  from  the 
steam,  to  40°. 

6.  Liquid  air  boils  at  (or  about)  — 182°  C.,  hence  all  the  heat  received 
by  the  liquid  air  at  this  temperature  is  used  to  convert  it  into  gaseous 
air. 

7.  The  good  conducting  metal  vessel  transmits  heat  from  the  ice, 
originally  at  0°  C.,  to  the  liquid  air  which  is  at  —182°,  far  below  the  tem- 
perature of  the  ice.    This  heat  from  the  ice  boils  the  liquid  air.    Finally 
the  ice  in  contact  with  the  vessel  reaches  approximately  — 182°  C.,  when 
the  action  nearly  ceases. 

8.  (1)  The  specific  heat  of  water  is  very  large. 

(2)  The  heat  of  fusion  and  the  heat  of  vaporization  are  also  very 
large,  hence  the  absence  of  water  means  the  absence  of  these  checks  to 
sudden  heating  and  cooling  of  the  air. 

9.  Water  vapor  into  liquid  water.    This  condensation  of  water  vapor 
means  the  generation  of  536  gm.  cal.  per  gm.  of  vapor  condensed  and  this 
heat  is  given  to  the  air. 

10.  The  condensation  of  the  steam  furnishes  the  heat. 


HEAT— Pages  221,  222 

1.  The  blanket  conducts  heat  poorly,  hence  it  keeps  in  the  heat  of 
the  human  body  and  equally  well  keeps  out  the  heat  of  the  warm  air 
from  the  ice. 

2.  The  statement  suggests  that  clothes  produce  more  or  less  heat 
instead  of  merely  keeping  in  the  heat  which  the  body  produces. 

3.  (1)  The  circulation  will  keep  the  temperature  very  nearly  uni- 
form. 

(2)  The  lower  parts  of  the  water,  because  they  receive  heat  so  rapidly, 
will  be  hotter. 

4.  Glass  is  a  very  poor  conductor  and  also  a  very  rigid  substance. 
A  sudden  change  in  the  temperature  of  one  part  of  a  thick  piece  of  glass 
causes  a  change  in  the  molecular  motion  in  that  part  while  the  condition 
of  the  rest  of  the  glass  is  not  changed.    The  strain  thus  produced  between 

MAN.  MUM.  PHYS. — 3 


34  MANUAL  FOR  MUMPER'S  PHYSICS 

the  parts  having  different  molecular  motion  is  often  sufficient  to  over- 
come cohesion  and  the  glass  breaks. 

5.  When  a  room  is  heated  by  means  of  hot  air  the  air  is  the  warmest. 
When  heated  by  steam  or  hot  water  the  air  is  also  warmest,  but  not  so 
much  warmer  as  in  the  case  of  hot  air.  When  heated  by  a  grate  fire  the 
air  is  cooler  than  furniture  and  other  solid  bodies. 

HEAT— Pages  229,  230 

1.  1  gm.  cal.  =  41900000   ergs. 

1  gm.  cm.  at  New  York=980.2  ergs. 

1  gm.  cal.  =  — 98Q  2     =  42?47  Sm-  cm. 
1  joule  =10000000  ergs. 

.      41900000 
1  gm.  cal.=  10000QQO=4.19  joules. 

2.  1  gm.  cal.  =  42747  gm.  cm.  of  mechanical  energy  to  produce  which 
a  body  having  a  weight  of  1  gm.  must  fall  42747  cm. 

Specific  heat  of  copper  is  .09.  To  warm  1  gm.  of  copper  5°  requires 
5 X. 09  =.45  gm.  cal. 

.45  gm.  cal.  =  .45X42747  gm.  cm.=  19236  gm.  cm. 
The  gm.  of  copper  must  fall  19236  cm. 

3.  It  lowers  the  temperature  since  some  of  the  heat  is  used  to  do  the 
work  of  expansion. 

4.  80X250=20000  Ib.  pressure  against  piston. 
20000X1^  =  30000  ft.  Ib.    Work  done. 

It  lowers  the  temperature  of  the  steam,  the  heat  energy  going  into 
the  mechanical  work. 

5.  It  means  that  10%  of  the  heat  developed  in  the  furnace  is  con- 
verted into  mechanical  work. 

6.  1  gm.  of  ice  requires  80  gm.  cal.  to  melt  it. 

1  gm.  eal.  =  42747  gm.  cm.    80X42747=3419760  gm.  cm. 
To  produce  this  the  ice  must  fall  3419760  cm. 
Any  mass  of  ice  must  fall  the  same  distance,  g  being  constant. 

7.  10%  of  7800=  780  gm.  cal.  of  heat  used  for  each  gm.  of  coal  burned. 
10  kilowatts  =  10000  watts. 

1  watt=  1  joule  per  sec.  =  3600  joules  per  hr. 

10  kilowatts  =3600X10000  joules  per  hr.  =  36000000  joules  per  hr. 

1  gm.  cal.  =  4. 19  joules. 


=  gm  Q  prouce         g  = 

gm.  of  coal  must  be  used. 

20  horse  power  =  15  kilowatts.  15  kilowatts  for  2  hr.  =  30  kilowatts 
for  1  hr.  Hence  three  times  as  much  coal  will  be  used  as  that  computed 
in  first  part  of  problem. 


SOUND  35 

SOUND—  Pages  256,  257,  258 

1.  The  vibrations  are  so  intense  and  the  pitch  of  some  of  them  so 
low  that  the  window  panes  are  able  to  take  up  the  vibrations.    They  are 
sympathetic  vibrations. 

2.  In  the  open  air  the  hearer  receives  only  the  direct  waves;  there  is 
little  or  no  reflection  as  it  occurs  in  rooms,  streets,  etc. 

3.  The  reflection  from  side  to  side  prevents  the  waves  from  spreading, 
hence  their  intensity  decreases  very  slowly  with  the  increase  in  distance 
from  the  source. 

4.  The  sound  waves  travel  to  the  woods  and  back,  a  total  distance 
of  \  mile  or  2640  ft.    Assuming  the  temperature  32°,  the  speed  is  1090  ft. 

2640 
per.  sec.,  hence  it  will  require  jQ9o=2.4+  sec. 

6.  The  speed  of  a  compressional  wave  (sound  wave)  varies  directly 
as  the  \/elasticity  and  inversely  as  the  \/density.  A  rise  in  the  tem- 
perature of  the  atmosphere  generally  decreases  the  density  without  chang- 
ing the  elasticity. 

6.  A  uniform  medium  undisturbed  by  currents  or  by  the  relative 
motion  of  its  parts  serves  the  purpose  best. 

7.  Speed  of  sound  waves  at  80°—  is  (48  ft.  per  sec.  faster  than  it  is  at 
32°),  1138  ft.  per  sec. 

In  2  sec.  the  waves  travel  from  the  source  to  the  hearer.  2X1138  = 
2276  ft. 

8.  See  text,  pages  231,  234. 

9.  If  5  sec.  are  required  for  the  waves  to  travel  to  the  hill  and  back, 
2.5  sec.  are  required  for  the  distance  to  the  hill.    Speed  is  1090+38=  1128 
ft.  per  sec.    In  2.5  sec.  the  distance  will  be  2.5X1128=2820  ft. 


10.   Wave  length=  =-         =  5'5  ft'    The  Period  is         of  a  sec" 


2>0 

11.  Wave    lengthXfrequency=  velocity.      4X256=1024    ft.    per    sec. 
Because  it  increases  the  speed,  by  decreasing  the  density,  it  will  increase 
the  wave  length. 

12.  An  octave  above  has  twice  the  frequency,  hence  half  the  wave 
length  of  the  fundamental.     A  change  of  the  medium  transmitting  the 
waves  of  a  given  note  changes  the  speed  and  wave  length  without  chang- 
ing the  pitch,  hence  the  relation  between  the  wave  lengths  gives  the  true 
relation  between  the  pitch  of  two  notes  only  when  the  waves  are  in  the 
same  medium. 

13.  Tones  having  same  pitch  have   same  frequency,   but  when   the 
media  are  different  they  may  have  different  velocities,  hence  different 
wave  lengths. 

«f  =725.+  per  sec.    Period  -^  sec. 
15.   Because  the  waves  are  constantly  enlarging  spherical  shells  there 


36  MANUAL  FOR  MUMPER'S  PHYSICS 

will  be  less  energy  per  unit,  surface,  even  though  we  do  not  count  the 
sound  energy  which  is  wasted  by  being  converted  into  heat  as  the  wave 
advances. 

16.  See  page  246,  lines  2-5. 

It  may  also  be  demonstrated  by  means  of  Fig.  266,  page  265.  The 
wasted  energy  makes  the  intensity  decrease  much  faster  than  it  would 
according  to  the  law  of  inverse  squares  alone.  Sound  waves  lose  their 
energy  more  rapidly  while  traveling  a  given  distance  in  sawdust  than 
they  do  when  traveling  the  same  distance  in  the  wood. 

17.  Mixed  materials,  especially  those  containing  a  great  deal  of  air,  are 
put  into  the  spaces  between  the  walls  and  floors,  etc.    These  substances 
transform  or  use  up  the  energy  of  sound  waves  very  rapidly. 

18.  They  reflect  the  waves  so  that  the  hearer  receives  them  along  with 
the  direct  waves,  thus  making  the  total  effect  or  intensity  greater  than 
it  would  otherwise  be. 

19.  Intensity  (energy  per  unit  surface  of  wave). 
Pitch  (frequency  of  vibration  of  particles  in  wave). 

Quality  (number  and  character  of  the  overtones  associated  with  the 
fundamental). 

20.  See  page  250,  sec.  177. 

21.  They  interfere  (254-250)  4  times  per  sec. 

22.  Law  2,  page  254. 

When  the  length  is  29  cm.  the  frequency  will  be  87  -f-  29 =3  times  the 
frequency  which  it  has  when  87  cm.  long,  other  conditions  being  the  same. 

23.  An  octave  higher  means  that  the  upper  note  will  have  twice  the 
frequency  of  the  lower,  to  produce  which  the  tension  must  be  made  4 
times  as  great.    4X41b.=  161b. 

Law  1,  page  254. 

24.  \  the  length  under  the  same  tension  would  give  a  note  an  octave 
higher.    To  reduce  the  pitch  back  to  the  original,  an  octave  lower,  the  ten- 
sion must  be  made  \  of  40  kil.=  10  kil. 

25.  Faster  at  noon  because  the  temperature  is  generally  higher  and 
the  density  of  the  air  less  then.    Faster  at  the  sea  level  because  generally 
warmer.    It  is  true  that  the  density  is  greater  at  sea  level,  but  as  the  elas- 
ticity is  correspondingly  greater,  they  neutralize. 

26.  The  compressed  air  has  both  a  greater  density  and  a  greater  elas- 
ticity, consequently  the  loss  in  one  respect  is  counteracted  by  a  gain  in 
the  other. 

LIGHT— Pages  269,  270 

1.  The  spherical  character  of  the  waves,  on  account  of  which  each 
point  on  a  wave  front  advances  along  a  straight  line,  thus  giving  straight 
rays. 


LIGHT  37 

.  See  Fig.  24. 
Let  bg  represent  4  ft.,  the  distance  of  the  screen  from  the  pin  hole  6, 

and  bf,  100  ft.,  the  distance  of  the  tree. 
Triangle  abc  is  similar  to  triangle  bde. 
abf  is  similar  to  beg  and  bcf  similar  to  bdg. 
Then  ac:  bf=de:  bg;  substituting, 
ac:  100=  £:  4. 
4ac=50. 
ac=12.5  ft. 


FIG.  24 


3.  When  moved  farther  from  the  pin  hole  the  image  becomes  larger 
and,  because  that  means  the  same  amount  of  light  distributed  over  more 
surface,  the  image  becomes  less  bright. 

4.  Light  acts  upon  the  organ  of  sight.    In  total  darkness  there  is  no 
light  to  do  the  acting.     In  blindness  there  is  not  a  proper  organ  to  re- 
ceive light  energy. 

6.  It  is  sharpest  at  the  end  of  the  shadow  nearest  the  pole.  See  page 
264.  The  bird  is  so  small,  compared  to  the  sun,  and  its  distance  from  the 
earth  so  great  that  its  true  shadow  or  umbra  does  not  extend  to  the 
earth.  See  page  264,  Fig.  265. 

6.  A  flame  is  seen  by  means  of  the  light  it  generates. 

7.  It  is  seen  by  its  own  light  when  the  current  is  on.     In  the  other 
case  it  is  seen  by  the  light  it  reflects;  and  being  nearly  black  it  reflects 
little  or  no  light  and  is  not  easily  seen. 

8.  Light  is  generated  by  the  flame  and  by  the  glowing  coal.     We 
cannot   see   the  ashes   unless   they   are  illuminated   from  some   outside 
source. 

9.  The  intensities  at  the  two  sources  are  to  each  other  directly  as  the 
squares  of  the  two  distances  from  the  surface  on  which  they  produce  equal 
illumination. 


38  MANUAL  FOR  MUMPER'S  PHYSICS 

Let  x=  candle  power  of  the  lamp,  then 
l:x=  (8)2;  (60)2 
=  64  :  3600 


z=56i-  c.  p. 

10.   Let  x=  distance  from  16  c.  p.  lamp. 
16:  75  =z2:  (450)2 

16  X  (450)2=3240000 


x=  207.8+  cm.,  distance  from  16  c.  p.  lamp. 

11.  See  Fig.  25.    When  held  in  the  position  ab  the  book  receives  all  the 
light  between  the  lines  Sa  and  Sb.    When  it  is  in  the  position  cd  it  receives 


FIG.  25 

only  the  light  between  the  lines  Sc  and  Sb.  The  source  has  the  same  in- 
tensity and  the  average  distance  of  the  book  from  the  course  is  the  same 
as  before. 

LIGHT— Pages  276,  277 

1.  Regular  reflection  occurs  at  a  smooth  or  practically  smooth  surface, 
essentially  all  the  light  going  off  from  each  point  in  only  one  direction. 
Irregular  reflection  occurs  at  a  rough  surface,  the  light  from  each  point 
being  sent  off  in  practically  all  directions.     A  point  as  here  used  really 
means  a  very  small  surface  and  not  a  mathematical  point. 

2.  It  could  not  be  seen  since  it  would  neither  diffuse  nor  absorb  any 
of  the  light  it  receives. 

3.  See  page  271,  par.  196. 

4.  When  looking  at  his  own  image,  the  distance  of  the  image  of  the 
observer  is  twice  the  distance  of  the  mirror  from  the  observer.    It  makes 
the  image  appear  smaller  and  in  consequence  a  person  can  see  his  entire 
image  in  a  mirror  £  his  height. 

5.  See  page  273,  sec.  198. 

6.  In  the  case  of  a  real  image  the  light  from  the  object  reaches  the 
place  where  the  image  is.    In  the  case  of  a  vertical  image  the  light  from 
the  object  does  not  reach  the  place  where  the  virtual  image  seems  to  be. 

7.  The  object  must  be  at  a  distance  from  the  mirror  which  exceeds 
the  principal  focal  length. 


LIGHT  39 

LIGHT— Page  282 

Index  of  glass       1.5240        ,.0.    .    ,        .. 

Index  of  water=  O34CT  L1424'  mdex  of  water  to  glass' 


2.  The  index  of  refraction  of  air  may  be  taken  as  1 : 

X  1860QO=  112650 ±  mi.  per  sec. 

=  139430  ±  mi.  per  sec. 

3.  See  page  280.     The  index  of  refraction  varies  with  the  speed  of 
light  in  the  medium. 

4.  It  is  refracted  more  when  going  from  air  into  a  glass  prism,  other 
things  being  the  same. 

1  524 
1  (nearl   )  =  l-^24>  index  of  refraction,  air  to  glass. 

1  524 
1  SS4()=  1-1424>  mdex  of  refraction,  water  to  glass. 

Hence  the  first  is  1.1424  times  the  second. 


LIGHT— Page  290 

1.  Drawings  based  on  sec.  204,  page  282. 

2.  The  convex  type.    The  difference  is  chiefly  in  the  brilliancy  of  the 
image  and  the  clearness  of  definition  or  of  outline. 

3.  *    U -i     1-1+-! 

/      di     d0'     12     20     do 

Clearing  of  fractions,  20d0=  12d0+240, 

d0=30  cm.,  dist.  of  object  from  the  lens. 

•Since  object  is  more  distant  than  image,  the  object  is  larger,  in  ratio 
of  30:  20—,  or  1^  times. 

4.  ~f=i^+l-     Clearing,  486=  9/+54/, 

63/=486, 

/=7.7  inches. 

6.   -.=  1+1. 


Which  means  that  the  image  is  also  twice  the  focal  length  distant 
from  the  lens.    The  image  and  object  are  consequently  of  the  same  size. 


40 


MANUAL  FOR  MUMPER'S  PHYSICS 


LIGHT— Page  302 

1.  Strictly  it  is  a  characteristic  of  the  sensation  due  to  the  kind  of 
light  which  the  object  sends  to  the  eye. 

2.  The  difference  which  produces  the  difference  in  color  effects  is  due 
to  the  difference  in  wave  lengths  or  in  the  vibration  frequency. 

3.  A  prism,  by  separating  the  light  into  its  constituents,  shows  that 
white  is  a  mixture  of  light  that  contains  essentially  all  the  wave  lengths 
which  act  upon  the  eye.     White  or  gray  color  sensations  may  be  pro- 
duced by  certain  combinations  of  two  or  more  colors  known  as  com- 
plementary, such  as  yellow  and  blue. 

4.  A  pure  black  means  a  complete  absence  of  light;  hence  there  is  no 
light  sensation  and  no  color.     Not  in  the  same  sense,  for  white  is  not  a 
color  but  a  combination  of  colors  or  color  effects. 

5.  In  sound  waves  a  similar  difference  would  produce  a  difference  in 
pitch. 

6.  It  depends  upon  the  kind  of  light  with  which  the  object  is  illumi- 
nated and  how  the  object  treats  the  light  it  receives;  that  is,  what  part 
it  transmits,  what  it  reflects  and  what  it  absorbs. 


MAGNETISM— Page  314 

1.  For  (a)  and  (6)  see  Fig.  322,  page  309;  for  C  see  Fig.  26. 

2.  See  Fig.  323.  The 
iron  armature  serves  best 
as  the  medium  to  com- 
plete the  magnetic  cir- 
cuit, so  that  the  poles 
of  the  magnet  cannot  be 
neutralized  or  reversed 
by  another  magnetic  in- 


N 


fluence  from  without. 


FIG.  26 

3.  Magnetic  induction  always  produces  an  opposite  pole  next  to  the 
inducing  pole,  and  opposite  poles  attract. 

4.  The  dipping  needle  shows  more  nearly  the  true  direction,  provided 
its  axis  is  at  right  angles  to  the  magnetic  meridian.    If  not  at  right  angles, 
the  dip  indicated  is  greater  than  the  true  dip. 

5.  Fig.  324,  page  309.    They  blend  when  the  poles  are  unlike. 

8X24     16 

6.  fa\->  =  "o~=5^.    The  poles  repel  with  a  force  of  5^  dynes. 


(6)2 


ELECTROSTATICS— Pages  329,  330 


1.   Positive  electrification  is  produced  on  the  glass  when  it  is  rubbed 
with  silk.     It  is  produced  whenever  any  two  different  substances  are 


CURRENT  ELECTRICITY  41 

rubbed  together,  but  this  is  the  most  convenient  way  of  identifying  it. 
The  name  positive  is  merely  intended  to  contrast  it  with  the  other  kind. 
The  names  might  have  been  reversed. 

2.  Repulsion  is  more  reliable,  for  a  body  electrified  either  way  will 
attract  an  unelectrified  body,  but  each  will  repel  only  the  kind  like  itself. 

3.  Let  us  call  the  electrified  bodies  A  and  B.     The  suspended  ball 
is  first  attracted  by  the  nearer  body,  say  A,  and  touches  it.    The  ball  shares 
the  charge  of  A,  is  repelled  by  A  and  attracted  by  B  until  it  touches  B, 
when  its  charge  is  reversed,  and  now  it  is  repelled  by  B  and  attracted 
by  A,  etc. 

4.  The  hand  and  body  being  relatively  good  conductors  remove  the 
electrification  as  fast  as  it  is  produced.    Support  the  rod  on  glass,  rubber 
or  silk. 

6.  See  sec.  236,  page  316. 

6.  The  positively  electrified  glass  rod  will  attract  the  negative  charge 
in  the  electroscope  and  the  leaves  will  come  together,  provided  the  glass 
rod  is  not  brought  so  near  as  to  produce  an  induced  charge. 

7.  A  further  divergence  of  the  leaves  shows  that  the  charge  in  the 
electroscope  was  repelled  by  that  on  the  sulphur.    From  this  it  is  evident 
that  the  sulphur  was  electrified  negatively,  the  same  as  the  electroscope. 

8.  The  charge  on  the  outside  is  distant  from  the  inside  charge  by 
just  the  thickness  of  the  glass  walls  of  the  jar.    Since  the  attraction  varies 
inversely  as  the  square  of  the  distance  between  the  charges,  neither  charge 
can  hold  the  other  completely  unless  they  are  at  the  same  place.    A  part 
of  each  charge  on  the  jar  is  consequently  not  bound  or  held  by  the  opposite 
charge.     This  excess  can  be  removed  by  alternately  touching  the  sides, 
this  always  leaving  an  excess  on  the  side  not  touched  last. 

CURRENT  ELECTRICITY— Pages  338,  339 

1.  A  containing  vessel,   a  fluid   (semifluid    or    paste)   and  two  con- 
ductors, the  latter  being  placed  in  contact  with  the  fluid  but  not  with 
each  other.    The  fluid  must  act  chemically  upon  only  one  of  these  plates, 
or  upon  one  more  than  upon  the  other. 

2.  See  page  332,  line  17.    Note  that  the  term  pole  has  a  very  different 
meaning  here  from  that  which  it  has  in  connection  with  magnets. 

3.  (a)  From  the  zinc  through  the  liquid  to  the  carbon.     (6)  From 
the  carbon  through  the  connecting  wire  to  the  zinc.     Both  are  the  direc- 
tions of  the  positive  discharge. 

4.  The  carbon  is  said  to  have  the  higher  and  the  zinc  the  lower  po- 
tential. 

5.  See  page  332,  line  21. 

6.  See  page  335,  sec.  254. 

7.  See  page  334,  sec.  253. 


42  MANUAL  FOR  MUMPER'S  PHYSICS 

8.  See  page  333,  lines  7-21.    If  a  circuit  is  left  open  when  not  in  use 
it  is  said  to  be  an  open  circuit  system,  but  if  kept  closed  when  not  in  use 
it  is  a  closed  circuit  system. 

9.  See  page  335,  sec.  255,  and  page  361,  sec.  275. 

10.  Uniform  materials  and  no  polarization.  High  E.  M.  F.  and  low 
resistance. 

CURRENT  ELECTRICITY— Pages  346,  347 

1.  A  charge  refers  to  an  electrical  condition  which  for  the  time  is 
fixed,  while  a  current  conveys  the  idea  of  a  continuous  discharging  or  flow- 
ing from  one  place  to  another. 

2.  By  sending  the  charge  through  a  coil  of  well  insulated  wire  which 
surrounds  a  piece  of  steel. 

3.  See  page  308,  line  8,  etc.     The  magnetic  field  can  be  shown  by 
means  of  a  magnetic  compass  placed  in  different  positions  with  reference 
to  the  conductor. 

4.  (a)  Toward  the  west.    (6)  Upward,    (c)  Toward  the  east,    (d)  Down- 
ward. 

6.  The  north  end  of  the  needle  is  deflected  toward  the  west. 

6.  The  current  flows  toward  the  south. 

7.  Clockwise  as  you  look  up  the  pole. 

8.  If  steel  is  used  the  magnet  made  is  permanent,  though  it  may  vary 
in  strength  with  a  variation  of  the  current. 

9.  From  the  observer.    Consult  Fig.  362,  page  342,  where  the  reverse 
is  shown. 

10.  The  direction  of  the  lines  of  force  in  the  magnetic  field,  produced 
by  the  second  and  additional  turns,  agrees  with  that  of  the  first  turn, 
hence  the  total  intensity  is  greater  when  the  number  of  turns  is  increased. 

CURRENT  ELECTRICITY— Page  358 

1.  See  page  354,  line  8,  etc. 

2.  The  metal  appears  at  the  cathode. 

3.  The  one,  the  cathode,  gains  as  much  as  the  other  loses. 

4.  Within  the  cell  the  current  is  said  to  flow  toward  the  plate,  gaining 
in  weight,  the  cathode. 

6.  Because  there  is  a  break  in  the  circuit  at  the  push  button,  which 
can  be  closed  only  by  pushing  the  button. 

6.  Because  the  vibrating  clapper  alternately  opens  and  closes  the 
circuit  at  the  bell,  thus  giving  a  succession  of  taps. 

7.  If  air  were  present  the  carbon  would  burn. 

8.  Platinum  is  the  only  metal  which  has  the  same  coefficient  of  ex- 
pansion as  that  of  glass. 

9.  Because  the  storage  cell  is  itself  "  charged  "  by  means  of  a  dynamo, 


CURRENT  ELECTRICITY  43 

and  because  of  the  necessary  loss  of  energy  in  the  transformation,  the  cell 
cannot  furnish  quite  so  much  energy  as  the  dynamo  puts  into  it.  Then  in 
addition  to  that  the  cost  of  the  cell  itself  must  be  included. 

10.  They  attract  each  other  because  they  are  parallel  and  flow  in  the 
same  direction  around  the  coil. 

CURRENT  ELECTRICITY— Page  363 

1.  6.5 -4-1000=. 0065  ohm  per  ft. 

88 X. 0065  =.572  ohm,  resistance  of  88  ft. 

2.  Since  iron  is  a  substance  with  less  conductance  (more  resistance), 
it  must  have  a  larger  cross  section.    It  must  be  6.63 -f- 1.08=  6.1  -f-  times  as 
large. 

3.  They  must  be  connected  end  to  end. 

They  must  be  connected  side  by  side,  that  is,  each  wire  must  have 
its  two  ends  connected  to  the  same  bodies  as  those  to  which  the  others 
are  connected. 

4.  Their  joint  resistance  is  3+6-j-9=  18  ohms. 
Their  joint  conductance  is  4+i-J-$=H  ohms. 

5.  The  resistance  varies  directly  as  the  length  and  inversely  as  the 
cross  section. 

Resistance=  1250  -=-106.3-^4=  2.93+  ohms. 

6.  It  decreases  the  length  and  hence  the  resistance  of  the  conducting 
liquid  between  the  plates,  hence  the  current  becomes  stronger. 

7.  Lifting  the  plates  when  they  are  at  a  constant  distance  decreases 
the  area  of  the  cross  section  of  the  conducting  liquid,  hence  increases  the 
resistance  in  the  circuit  and  the  current  is  weakened. 

CURRENT  ELECTRICITY— Page  368 

1.  A  water  current  is  due  to  a  difference  between  the  pressures  at  the 
two  points. 

An  electric  current  is  due  to  a  difference  between  the  potentials  of  the 
two  points. 

There  may  be  no  current  in  either  case  if  the  medium  between  the  two 
points  is  not  a  sufficiently  good  conductor,  that  is,  if  it  offers  too  much 
resistance  to  the  flow. 

2.  The  so-called  current  strength  is  determined  by  the  quantity  of 
electricity  transferred  per  unit  of  time  and  not  the  energy  of  this  elec- 
tricity, just  as  the  strength  of  a  water  current  depends  upon  the  quantity 
of  water  flowing  and  not  upon  its  energy. 

3.  The  unit  current  strength  is  defined  in  terms  of  the  chemical  effect 
produced.    See  page  359,  sec.  274. 

4.  The  current  strength  increases  with  the  increase  of  the  conduct- 
ance (decrease  of  resistance)  of  the  conducting  body  and  vice  versa. 


44  MANUAL  FOR  MUMPER'S  PHYSICS 

5.  Page  362,   sec.  276.     The  substance  will  have  -3-  of  a  unit  con- 
ductance. 

6.  See  page  365,  sec.  278.     See  page  367,  sec.  281. 

7.  It  would  furnish  more  than  1  ampere  of  current. 

8.  It  would  produce  more  than  1  ampere  through  1  ohm  resistance. 
The  power  would  likewise  be  more  than  1  watt. 

9.  See  page  367,  sec.  281. 

10.  See  page  129,  sec.  76.' 

11.  (a)  Po  wer  =  volts  X  amp.  =  60X10  =  600  watts. 
(6)  600  watts  would  furnish  600  candle  power. 

(c)   600  watts  =^=.6  kilo  watt. 

.6  kilowatt,  at  6  cents  per  kilowatt  hour,  give  .6X6  cents  = 
3.6  cents  per  hour. 

12.  H=.24C2Rt. 

gm.  cal.=  .24X  (12)*X  1.5X300=  15552. 

CURRENT  ELECTRICITY—  Page  375 

1.  See  text. 

volts      6 


3.  Volts=  amp.  Xohm=  10X20=  120  volts. 

volts     110     __„     , 

4.  Ohms  =       —  =  —    =  220  ohms. 


5.  Volts=amp.Xohms=12X.8=9.6  volts. 

6.  1  amp.  sets  free  .001118  gm.  per  sec. 
10  min=600  sec. 

1  amp.  in  600  sec.  sets  free  600X.  001118  gm.=  .6708  gm.  of  silver. 
6  amp.  set  free  6  X.  6708  =4.0248  gm.  of  silver  in  10  min. 

volts=  amp.  Xohms=  6X24  =144  volts. 

7.  Resistance  =50  ohms. 
Amp.Xohms=volts=.2X50=10  volts. 
To  produce  10  volts  it  will  require 
f$=6$;  seven  cells  will  be  required. 

Because   external   resistance  is  large  the   cells  are  arranged  in 
series. 

8.  6X2=12  volts,  E.  M.  F. 

6X1.5  ohms  =9  ohms,  battery  resistance. 

9.  When  abreast  the  total  E.  M.  F.  is  the  same  as  that  of  a  single  cell, 
in  this  case  2  volts.    The  resistance  is  £  that  of  a  single  cell. 

£  of  1.5  ohms=  .25  ohms. 
10.  See  page  370. 
(1)   For  external  resistance  of  90  ohms. 


CURRENT  ELECTRICITY  45 


Series,  Am.=  -  .  —  ™^=      =  -121  amP- 

ex.  ohms  4-inter  .  ohms     90  +  (6  X  1  .5)     99 


The  series  current  is  nearly  6  times  as  strong  as  the  parallel. 
(2)   For  external  resistance  of  .2  ohms. 

1O 


Paralle1'  .2+KT5)==4-4  amp' 

The  parallel  arrangement  gives  nearly  4  times  the  current. 
11.  Power  in  watts  =  amp.  X  volts.    See  page  367. 

Volts  =  amp  .  X  ohms,  hence  watts  =  amp  .  X  amp  .  X  ohms  . 

Power  required  for  each  lamp=  (.4)2X60=.  16X60=  9.6  watts. 

^=520  lamps. 
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